Saturday

April 19, 2014

April 19, 2014

Number of results: 1,494

**plz integrate**

assume you mean (1+u) du /(3u+2u^2) which is (1+u) du/[u(3+2u)] which is 1 du/[u(3+2u)] + u du/[u(3+2u)] now let's see if we can expand 1/[u(3+2u)] into a/u + b/(3+2u) = 1/ [u(3+2u)] a(3+2u) + b(u) = 1 3 a = 1 so a = 1/3 2a + b = 0 so b = -2/3 so we have 1/[u(3+2u)] = (1/3)/u...
*Wednesday, October 8, 2008 at 7:22pm by Damon*

**calculus**

let u = y + 5 then du = dy we have integral of (u-5) u^(1/3) du or of u^(4/3) du - 5 u^(1/3) du Can you take it from there? Remember to change u back to y+5 in the end and add a constant c.
*Saturday, May 14, 2011 at 7:25pm by Damon*

**Calculus 2 correction**

I have u = x^2 + 2 so x^2 = (u-2) du = 2 x dx so x dx =(1/2) du split into three integrals 7 x^3 dx/u + 2 x dx/u - 3 dx/(x^2+2) 7x^2 xdx/u+du/u-(3/sqrt2)tan^-1(x/sqrt2) (7/2)(u-2)du/u +du/u - last term (7/2)du -7du/u+du/u - last term (7/2)u - 6du/u - last term (7/2)(x^2+2) - ...
*Friday, April 27, 2012 at 8:14pm by Damon*

**Math**

integrate (x+1)sqrt(2-x)dx u = (sqrt(2 - x)) u^2 = 2 - x x = 2 - u^2 - 2u du = dx I = integral sign I (2 - u^2 + 1)u -2u du I (3 - u^2)-2u^2 du I (-6u^2 + 2u^4) du -6 I u^2 du + 2 I u^4 du 2 I u^4 du - 6 I u^2 du 2 (1/5 u^5) - 6 (1/3 u^3) 2/5 u^5 - 2u^3 substitute back in for ...
*Monday, January 17, 2011 at 6:08pm by helper*

**Calculus**

solve for dv/du dQ/du= 38u + [(u+v)^2(18+19dv/du) -(2(18u+19v)(u+v)(1+dv/du)]/ (u+v)^4 u=10 v=25 I am confused how u solve for the dv/du in this big equation and reducing and stuff like that thanks!! this is about implicit di
*Monday, October 7, 2013 at 9:05am by Rebekah*

**Math differentiation+substitution**

We can solve this by differentiation of a function of a function. Given g(u) = (5+u^2)^5(3-9u^2)^8 let p(u)=(5+u²), and q(u)=(3-9u²) Then g(u)=p(u)^5 * q(u)^8 Using the chain rule, we get d(p(u)^5)/du =5p(u)^4*dp(u)/du =5p(u)^4*2u ...(1) Similarly, d(q(u)^8)/du =8q(u...
*Monday, February 21, 2011 at 11:15pm by MathMate*

**calculus**

again let u = (y+5) then du = dy and y = u-5 y^3 = u^3-15u^2+75u-125 and integral of (u^3-15u^2+75u-125)u^(1/2) du or u^(7/2) du - 15 u^(5/2)du + 75 u^(3/2) du - 125 u^(1/2) du which is (2/9)u^(9/2) ..... etc
*Saturday, May 14, 2011 at 7:20pm by Damon*

**Calculus**

That is a fine substitution to make u = x + 1 dx = du x^2 = u^2 - 2u+1 so we have integral (u^2-2u+1)(u^.5) du that is integral of u^2.5 du - integral of 2u^1.5 du + integral of u^.5 du in general integral of u^n du = [u^(n+1)]/(n+1)
*Sunday, December 9, 2007 at 3:26pm by Damon*

**integral confusion**

Well, dx= du/(2sec2x tan 2x) from your equation. INT 2sec2x tan2x * dx= INT du INT du= u On the second. You have u^30 du/3 since dx=du/3 INT u^30 du/3 = 1/3 INT u^30 du= 1/3 * 31 u^31 then on both, substitute back in the f(x) for u.
*Monday, November 26, 2007 at 11:21pm by bobpursley*

**Math integrals**

What is the indefinite integral of ∫ [sin (π/x)]/ x^2] dx ? This is what I did: Let u = π/x (to get the derivative and du:) π*1/x π(-1/x^2)dx = du π(1/x^2)dx = (-1)du so, 1/x^2 = -1/πdu then ∫ [sin (π/x)]/ x^2] dx = ∫ sin(u...
*Friday, August 15, 2008 at 1:31am by Marissa*

**Calculus PLEASE HELP**

r = Revenue c = cost p = profit = r-c u = number of units sold dr/du = marginal revenue dc/du = marginal cost so dp/du = dr/du - dc/du = marginal profit at peak profit, changer in profit for another unit sold is zero so 0 = dr/du-dc/du or dr/du = dc/du
*Thursday, December 12, 2013 at 5:29am by Damon*

**Calculus **

integrate x/(x^4+x^2+1) | = integrate symbol u = x^2 du = 2x dx 1/2 du = x dx 1/2 | du/(u^2 + u + 1) Complete the square u^2 + u = -1 u^2 + u + 1/4 = -1 + 1/4 (u + 1/2)^2 + 3/4 1/2 | du/((u + 1/2)^2 + 3/4) w = u + 1/2 dw = du 1/2 | dw/(w^2 + 3/4) 1/2 | dw/(3/4 + w^2) And, ...
*Saturday, January 22, 2011 at 8:11pm by helper*

**Calculus**

which of the following integrals results from making the substitution u=x^3 in orer to find (squiggly vertical line)x^2cos(x^3)dx ~cos u du ~u^2 cos u du ~u^(2/3) cos u du1/3 os u du ~3 cos u du
*Friday, May 11, 2012 at 7:11pm by Ken*

**Calculus**

if you let u = t^3+4 then du = 3t^2 dt Now your integrand becomes u^-2 (du/3) = 1/3 u^-2 du that should be a cinch.
*Thursday, April 25, 2013 at 11:19am by Steve*

**Calculus**

Thanks for help on 1. on 2. if i do u=sqrt(x) my du is 1/2x^(-1/2) and that means my du is in the denominator. So it would read 2integral of sin(u)/du
*Monday, April 11, 2011 at 10:25pm by Leanna*

**calculus**

find the partial derivatives du/dx, du/dy and du/dt. u= root (r^2+s^2, r = y + x cos(t), s = x + y sin(t);when x = 0, y = 2, t = 2
*Monday, October 10, 2011 at 1:47am by Ramesh*

**Calculus**

look into trig substitutions let y = 2sinu then 4 - y^2 = 4 - 4sin^2 u = 4 cos^2 u and √(4 - y^2) = 2cos u dy = 2 cosu du and your integral is just 1/(2 cosu) * 2 cosu du = 1/2 du integral(1/2) du = u/2 = (1/2) arcsin(y/2) + C
*Saturday, January 7, 2012 at 4:52pm by Steve*

**calculus**

Let 2x -7 = u so dx = du/2 Integrand becomes 2/u^2*(du/2) = du/u^2 from u = 1 to 5. = -1/u @ u = 5 - (-1/u @ u = 1) = 1 - 1/5 = 4/5
*Friday, June 24, 2011 at 11:15am by drwls*

**Calculus**

Find the derivative: y=2e^(x^2+1)^3 i have so far 2(2x)e^(x^2+1)^3 Is this correct? How do I simplify furthere if it is? Hmmmm. d/dx (f(x))= d/du(f(u)) du/dx let u= (x^2+1)^3 y= 2e^u dy/du= 2e^u du/dx= 3*(x^2+1)^2 * 2x so... dy/dx= dy/du *( du/dx)
*Tuesday, April 17, 2007 at 4:36pm by Evan*

**Calc.**

Evaluate the indefinite integral. Please check my work? Not sure if I am doing this correctly. S= integral symbol S x^3*sqrt(x^2 + 1) dx u = x^2+1 x^2= u - 1 du = 2xdx du/2 = xdx 1/2 S x^2 * x sqrt (x^2 + 1) du 1/2 S (u - 1) * sqrt (u) du multiply square root of u and (u - 1) ...
*Thursday, January 27, 2011 at 11:15pm by CMM*

**Calculus**

I think you may be going about things backwards. If u = 3/x du = -3/x^2 dx and your integrand becomes e^u du/(-3) = -1/3 Int(e^u du) = -1/3 e^u = -1/3 e^3/x check: taking the derivative of -1/3 e^3/x we get -1/3 * e^3/x * -3/x^2 = e^(3/x) / x^2 So, evaluating at 3 and 1, -1/3 ...
*Monday, November 21, 2011 at 12:12am by Steve*

**calculus**

1) y'=d/dt (uv^-1) where u= t^2-1 du=2dt and v=(t^2+2)^3 dv= 3(t^2+2)^2 * (2t)=6t(t^2+2)^2 y'= v^-1 du -uv^-2 dv y'=du/v -udv/v^2 then put in u,v du, dv and you have it.
*Sunday, June 24, 2012 at 7:54pm by bobpursley*

**calc**

correction: 9/25(u+7)u^11 du = 9/25(u^12 + 7 u^11) du and you can easily integrate this indefinate integral of 9x(5x-7)^11 dx using the u substitution method: i know u= (5x-7) but when i use this the x in the 9x doesn't cancel thus x= (u+7)/5. however when i put this in the ...
*Monday, December 4, 2006 at 2:18pm by Count Iblis*

**calculus**

yes. the substitution is correct: let u = x^(1/2) thus du = 1/[2(x^(1/2))] dx, or dx = 2(x^(1/2)) du, or dx = 2u du substituting these to original integral, integral of [e^x^(1/2) / x^(1/2)] dx integral of [(e^u) / u] * (2u) du the u's will cancel out: integral of [2*e^u] du ...
*Monday, May 2, 2011 at 10:41pm by Jai*

**math**

let u= sqrt(t) du= 1/2sqrt(t) dt dt= 2u du int e^-u * 1/u * 2u du=2 int e^-u du = -2e^-u eval 0 to inf= 2 check my thinking.
*Friday, April 30, 2010 at 9:49am by bobpursley*

**College Calculus 1**

d/dx((6-x)^(1/3) x^(2/3)) | Use the product rule, d/dx(u v) = v ( du)/( dx)+u ( dv)/( dx), where u = (6-x)^(1/3) and v = x^(2/3): = | x^(2/3) (d/dx((6-x)^(1/3)))+(6-x)^(1/3) (d/dx(x^(2/3))) | Use the chain rule, d/dx((6-x)^(1/3)) = ( du^(1/3))/( du) ( du)/( dx), where u = 6-x ...
*Thursday, October 14, 2010 at 3:16pm by Anonymous*

**Calculus**

Booo! 1/(a+b+c) is NOT 1/a + 1/b + 1/c What you need to do is let u = 3x+5 and you have du = 3 dx, so dx = du/3 x in [-1,2] means u in [2,11] giving you ∫[2,11] -4/u^2 du/3 -4/3 ∫[2,11] u^-2 du That should be ever so simple.
*Friday, August 10, 2012 at 2:02pm by Steve*

**math**

integral x/(x+5)^1/2 (u-5)/u^1/2 du giving u^1/2 du -5u^-1/2 du and integral will be now 2/3 u^3/2 -10u^1/2 +c =2/3(x+5)^3/2 -10(x+5)^1/2 +c =2/3(x+5)^1/2{(x+5) -30} +c =2/3(x+5)^1/2{(x -25} +c have i done something wrong??
*Thursday, November 19, 2009 at 6:31pm by atif*

**calculus**

Let u^2 = 3x-1 2u du = 3 dx and the integrand then becomes (u^2+1)/u * 2u/3 du = 2/3 (u^2+1) du Integrate that and then change back to x.
*Thursday, February 23, 2012 at 3:34am by Steve*

**Calculus**

Integral of x / sqrt(x-1) dx We can use substitution method. Let u = x - 1 Thus, x = u + 1 and dx = du Substituting, Integral of (u+1)/sqrt(u) du Integral of (u+1)(u^(-1/2)) du Integral of u^(1/2) + u^(-1/2) du = (2/3)*u^(3/2) + (2)*u^(1/2) + C = (2/3)*(x-1)^(3/2) + 2(x-1)^(1/...
*Wednesday, November 27, 2013 at 7:22am by Jai*

**calculus**

hmmm. Doesn't look too bad. u = 1-√x x = (1-u)^2 dx = -2(1-u) du dx/(x(1-√x)^2) = -2(1-u) du / ((1-u)^2 * u^2) = -2 /((1-u)*u^2) du = -2/u^2 -2/u + 2/(u-1) du = 2(1/u + ln(1-u) - ln(u)) Now substitute u = 1-√x to get back to x
*Friday, September 14, 2012 at 1:20am by Steve*

**Calc--derivatives**

I will try with an example. Use the function of a function rule, namely dy/dx = dy/dy . du/dx substitute u=log(x) and find the derivative of d(x^u)/du then multiply by du/dx. Suppose y=sin(x²) We know how to differentiate sin(x), and how to do x², but we're not sure ...
*Tuesday, October 13, 2009 at 5:45pm by MathMate*

**Calculus - derivatives**

let u=xx ln(u) = x ln(x) (1/u)du/dx = ln(x) + 1 du/dx = u (ln(x) + 1) du/dx = xx(ln(x) + 1) So this confirms what you've got for d(xx)/dx To find (x^x)^(x^x), it is not as complicated as it looks once we've got the derivative of xx. I would start with u=xx and du/dx = xx(ln(x...
*Sunday, November 29, 2009 at 3:58am by MathMate*

**Physics 102 (College)**

I'm stuck on part B of this question: A system does 125 J of work on environment and gains 79 J of heat in the process. Find change in internal energy (delta U, abbreviated dU)of a. the system and b. the environment. A. Find dU of the system W = 125 J Q = 79 J dU = Q - W dU...
*Sunday, February 27, 2011 at 2:46pm by Carla*

**Physics 102 (College)**

I'm stuck on part B of this question: A system does 125 J of work on environment and gains 79 J of heat in the process. Find change in internal energy (delta U, abbreviated dU)of a. the system and b. the environment. A. Find dU of the system W = 125 J Q = 79 J dU = Q - W dU...
*Sunday, February 27, 2011 at 3:20pm by Carla*

**elements of calculus **

d"=x/u where u = x^2+1 d"'= 1/u - x/u^2 * du/dx d""= 1/u^2 -1/u^2 du/dx +2x/u^3 (du/dx)^2- x/u^2 d"u/dx" so now it is the algebra du/dx= 2x d"u/dx"= 2 go for it. check my work.
*Tuesday, October 11, 2011 at 3:59pm by bobpursley*

**Calc. Checking Answer**

Find the antiderivative by hand in each case. S stands for the integral sign I want to make sure I am doing these correctly. A) S x*sqrt(10 + x^2) dx So, u= 10 + x^2 du= 2xdx du/2= xdx (1/2) S sqrt(u) du (1/2)*((u^(3/2))/(3/2)) (1/2)*(2/3)*(u^(3/2)) (1/3)*(u^(3/2)) = (1/3)*(10...
*Wednesday, April 6, 2011 at 5:36pm by Chelsea*

**math**

let u = (2t+1) du = 2 dt so dt = .5 du so .5 integral du sin u/cos^2u try 1/cos u d/du (1/cos u) = -sin u/cos^2 u You can take it from there :)
*Sunday, February 7, 2010 at 4:23pm by Damon*

**Calc 1**

if we let u = 1+7tanx du = 7sec^2 x dx and you have ∫ u^(-1/3) 1/7 du if we let u = sin√x du = cos√x * 1/(2√x) and you have ∫ 1/√u 2 du
*Friday, November 30, 2012 at 2:34pm by Steve*

**Calculus (Checking Answer)**

Find the antiderivative by hand in each case. S stands for the integral sign A) S x*sqrt(10 + x^2) dx So, u= 10 + x^2 du= 2xdx du/2= xdx (1/2) S sqrt(u) du (1/2)*((u^(3/2))/(3/2)) (1/2)*(2/3)*(u^(3/2)) (1/3)*(u^(3/2)) = (1/3)*(10 + x^2)^(3/2) correct/incorrect? B) S (x/(sqrt(2...
*Tuesday, April 5, 2011 at 5:07pm by Chelsea*

**Calc**

Let u = 1+ln x so, du = 1/x dx ln x = u-1 So, we have u^1/2 * (u-1) du = u^3/2 - u^1/2 du Integrate that to get 2/5 u^5/2 - 2/3 u^3/2 = 2/5 (1 + ln x)^5/2 - 2/3 (1+ln x)^3/2 Check your work by taking the derivative. Trust me -- it comes out right.
*Wednesday, October 19, 2011 at 1:05am by Steve*

**calculus**

use substitution u=1+x^4 du = 4x^3dx du/4 = x^3dx so ∫x^3(1+x^4)^6dx =(1/4)∫du/u^6 Use the power rule to integrate and evaluate between limits.
*Saturday, April 30, 2011 at 9:27pm by MathMate*

**Calculus AP **

integral of [sqrt(u) - 2 u^2]/u du i got stuck ∫u^-1 (sqrt(u) - 2 u^2) du ∫ u^(-1/2) - 2u du i dont know i'm doing right step
*Sunday, August 26, 2012 at 6:53pm by Vicky*

**Integral Help**

Most of these can be solved by the method of variable substitution. Define one function of x to be "u" and then calculate du. For example, in #3, let u = x^2 so that du = 2x dx. Then ç xe^x^2 dx = (1/2)ç e^u du = (1/2) e^u = (1/2) e^(x^2) See how many of the others you can ...
*Thursday, May 5, 2011 at 4:45am by drwls*

**calculus**

let u = e^x+1. du = e^x dx and you have ∫(u-1)/u du = ∫ 1 - 1/u du = u - lnu = (e^x+1) - ln(e^x+1) + C by absorbing the +1 into the C, you end up with = e^x - ln(e^x+1) + C
*Thursday, October 4, 2012 at 11:24am by Steve*

**calculus**

let u = 2x-1 x = (u+1)/2 du = 2 dx dx = du/2 x*sqrt(2x-1) = (u+1)/2 * sqrt(u) * du/2 = 1/4 * (u^3/2 + u^1/2) du can you take it from here?
*Thursday, February 23, 2012 at 11:35pm by Steve*

**Calculus**

Derivative Differentiate f(x)=(5x-4)^2 There are two ways to do this: (1) let u = 5x-4, and use the "chain rule", df/dx = df/du * du/dx OR (2) expand the polynomial to f(x) = 25 x^2 -40x + 16 and differentialte it term-by-term. You will get the same answer either way. Let's ...
*Monday, February 26, 2007 at 10:52am by Dee*

**Calculus**

Try substitution: u=4t du=4dt, or dt=(1/4)du then ∫f(4t)dt [0,0.25] =∫f(u)(1/4)du [0,4*0.25] =(1/4)∫f(u)du [0,1] =(1/4)*11 =11/4 The other problems should work similarly.
*Monday, September 5, 2011 at 2:48am by MathMate*

**Calculus Test Question!!!!!**

Look up the derivative of arc-secant. If y = sec^-1 u, dy/du = 1/[u sqrt(u^2-1)] Then let x+3 = u and use the chain rule. dy/dx = dy/du * du/dx
*Saturday, October 6, 2007 at 10:01pm by drwls*

**Calculus**

f(x)=4x sqrt(1-x²) use substitution u=1-x^2 du=-2xdx ∫f(x)dx =∫(4xsqrt(u))du/(-2x) =-∫2sqrt(u)du =-(4/3)u^(3/2)+C =-(4/3)sqrt(1-x^2)^(3/2)+C
*Friday, May 11, 2012 at 7:17pm by MathMate*

**Calculus**

a. This just a simple power rule: ∫x^n dx = 1/(n+1) x^(n+1) b. same here, also ∫e^u du = e^u Let u=2x so du = 2 dx c. Recall that ∫ 1/u du = ln(u) Let u = x+3 Don't forget the +C on indefinite integrals
*Saturday, April 27, 2013 at 1:40pm by Steve*

**Math please help urgent !**

dy/dx = dy/du du/dx du/dx = (1/3)x^-(2/3) dy/du = [(u-1)-(u+3)]/(u-1)^2 = [(x^(1/3)-1)-(x^(1/3)+3)/(x^(1/3)-1)^2 = -4 / (x^(2/3) -2x^(1/3)+1) so dy/du * du/dx =(-4/3)(x^-(2/3)) /(x^(2/3) -2x^(1/3)+1) = -4/3(x^4/3-2x +x^(2/3))
*Thursday, March 1, 2012 at 5:53pm by Damon*

**math**

Let x/8 = u dx = 8 du So you want the integral of 8*sin^2 u du sin^2 u = (1/2)[1 - cos(2u)] => integral of 4[1 - cos(2u)] du Let w = 2u; du = dw/2 = 4u - 2 sin(w) = x/2 - 2 sin(2u) = x/2 - 2sin(x/4) Check my thinking. I tend to be sloppy
*Wednesday, June 22, 2011 at 7:49am by drwls*

**Calculus**

Why plural? There is only one derivative of that function. Let x+1 = u, so that m(x) = m{u(x)}, m(u) = 1/u, and du/dx = 1 Then use the chain rule dm/dx = dm/du*du/dx = -1/u^2 = -1/(x+1)^2
*Friday, September 12, 2008 at 9:31pm by drwls*

**college math**

if u = x^2 then du = 2 x dx if dv = e^-x dx then v = -e^-x u v = -x^2e^-x v du = -2 x e^-x dx u v - integral v du = -x^2 e^-x + 2 integral x e^-x dx now do integral x e^-x dx u = x so du = dx dv = e^-x dx so v = -e^-x u v - v du = -xe^-x -integral -e^-x dx = -x e^-x - e^-x put...
*Monday, March 9, 2009 at 11:31pm by Damon*

**calculus**

Put tan(x) = u ----> x = arctan(u) ----> dx = du/(1+u^2) Integral of u^6/(1+u^2) du = Integral of (u^6 + u^4 - u^4 - u^2 + u^2 + 1 -1)/ (u^2 + 1) du = Integral of (u^4 - u^2 + 1) du - Integral of 1/(1+u^2) du = 1/5 u^5 - 1/3 u^3 + u - arctan(u) + c= 1/5 tan^5(x) - 1/3 ...
*Sunday, June 24, 2012 at 2:12pm by Count Iblis*

**Calculus URGENT test tonight**

Integral of: __1__ (sqrt(x)+1)^2 dx The answer is: 2ln abs(1+sqrt(x)) + 2(1+sqrt(X))^-1 +c I have no clue why that is! Please help. I used substitution and made u= sqrt(x)+1 but i don't know what happened along the way! Your first step was a good one. next, let u = sqrt x + 1 ...
*Friday, April 20, 2007 at 3:59pm by Gabriela*

**Calculus BC**

You'll need the chain rule, which says dy/dx = dy/du . du/dx Set 5ln(x) as u, then you proceed with f'(x) = 2cos(u).du/dx noting that d(ln(x))/dx = 1/x Can you take it from here?
*Thursday, October 29, 2009 at 2:20am by MathMate*

**Chain Rule**

2) Let u = 3x+5 and use the chain rule dy/dx = dy/du*du/dx y = u^8 dy/du = 8 u^7 u = 3x+5 du/dx = 3 You complete the problem. If you need help with the others, show your work and help will be provided.
*Wednesday, September 26, 2007 at 5:47pm by drwls*

**Calculus**

You memorize the standard forms: In this case if y=a x^b where b is any +- number dy/dx= b*a* xb-1 here is an example: y= 1/sqrt(x^2+1) y= (x^2+1)^-.5 let u= x^2+1 dy/du= -.5 (u)^-1/5 dy/dx= dy/du*du/dx= -.5(x^2+1)^-1.5 * 2x because du/dx= d (x^2+1)/dx= 2x
*Tuesday, February 28, 2012 at 5:41pm by bobpursley*

**Math(Please check)**

evaluate the integral integral of 3 to 2 x/(x^2-2)^2 dx u=x^2-2 du=2x dx 1/2 du = x dx integral of 1/u^2 du -1/(x^2-2) Then I plug in 3 and 2 and subtract them form each other -1/(3^2-2) - (-1/(2^2-2) Is this correct?
*Saturday, April 30, 2011 at 6:06pm by Hannah*

**applied mathmatics**

dy/du = 6u - 6 du/dv = 2v dv/dx = 2 dy/dx = (dy/du)(du/dv)(dv/dx) = (6u-6)(2v)(2) = 24v(u-1) If necessary, use substitution of the originals to express in terms of whichever variable you want
*Friday, May 25, 2012 at 4:20am by Reiny*

**Math - Derivatives**

d/dx e^u = e^u du/dx here u = x^-1 du/dx = -1 x^-2 so d/dx e^(x^-1) = -x^-2 e^(x^-1) = -(1/x^2)e^(1/x^2) Try the others using that recipe: d/dx e^u = e^u du/dx
*Wednesday, January 28, 2009 at 3:49pm by Damon*

**college math**

Remember the "chain rule". Let u(x) = x^2 du/dx = 2x d e^(x^2)/dx = (de^u/du)(du/dx) = 2x e^(x^2) =
*Thursday, February 26, 2009 at 5:05pm by drwls*

**Brief Calculus**

Let u = 2x^2 -1 Then du = 4x dx, and x*dx = du/4, x/(2x^2 − 1)^0.4 dx = (du/4)*u^-0.4 You can easily integrate that.
*Monday, May 2, 2011 at 4:57am by drwls*

**Calculus**

this is a classic case for trig substitution Let x = 2secu Then x^2 - 4 = 4sec^2 u - 4 = 4tan^2 u dx = 2 secu tanu du 1/(x^2-4)^1/2 dx = 1/(2tan u) * 2 secu tanu du = secu du Now, Int(secu du) = ln(secu + tanu) = ln(x + sqrt(x^2 - 4)) + C question: why can it be this, instead ...
*Sunday, December 11, 2011 at 3:23pm by Steve*

**math**

∫sin√x dx let u = √x du = 1/(2√x) dx dx = 2√x du = 2u du ∫sin√x dx = ∫sinu (2u du) = 2∫u sinu du now use integration by parts to get 2sinu - 2u cosu and convert back to x's
*Friday, October 12, 2012 at 11:38am by Steve*

**Calculus**

Use the fact that , for f(u) = arcsin u, df/du = 1 / sqrt(1 - u^2) Use the chain rule: df/dx = df/du* du/dx Let u = 3x, so du/dx = 3 df/dx = 3 / sqrt(1 - 9x^2)
*Friday, October 5, 2007 at 9:55pm by drwls*

**Calculus AP**

Use the table of integrals to find int cos^4 3x dx I found the table: ∫cos^n u du = (1/n)cos^(n-1)u sinu + (n-1/n)∫sin^(n-2)u du = 1/4 cos^(4-1)u sinu + (4-1/4)∫sin^(4-2) u du so what i did the problem: let u=3x then du=3dx =1/4*1/3 cos^3u sinu + 3/4*1/3 &#...
*Thursday, September 20, 2012 at 1:42am by Vicky*

**calculus**

Well, I happen to know that the derivate of the tangent is sec^2 :) (cosx)/(sin^3x) dx Try f(x) = 1/sin^2 x f' = - 2 sin x cos x /sin^4 x see what happens? let u = 2 x then du = 2 dx and we have csc u cot u du/2 from u = pi/2 to pi/6 (1/sin u)(cos u/sin u) du/2 = (1/2) (cos u/...
*Sunday, January 6, 2008 at 6:09pm by Damon*

**Calculus**

1) dy/dx = dy/du*du/dx = (3 u^2 + 6)(8x^3 + 6x) To get it in terms of x only, substitute the u(x) function for u. 2) Use the same "function of a function" approach as above. Let u = 4x^2 -x f(u) = u^(-1/2) df/dx = df/du*du/dx 3) Let u = (2x-5) f(u) = 5/u^(2/3) Use the same ...
*Wednesday, May 21, 2008 at 7:37am by drwls*

**Integral Calculus**

To solve that last integral, you have to use the substitution rule. u= x^2 + 1 du= 2x Therefore the integral becomes: -1/2∫(2/u) du Factor out the 2 -∫1/u du Integrate -ln(u) Which is equal to -ln(x^2 +1) Since u= x^2 + 1
*Thursday, May 24, 2012 at 10:29pm by Ethan*

**French continued**

1) ....je vais au magasin 3.4.5.)Even though there is a "de la", there is no "de le". It becomes "DU" (not "du le") 5)Je rentre du café et je vais à la maison. 3)Je rentre du collège et de la papeterie. 4)Je rentre du magasin et de la Fête de Dominique.
*Monday, March 2, 2009 at 4:07pm by E.G*

**Calculus**

Let u^2 = x+5 2u du =dx x = u^2 - 5 x^2 - 3 = u^4 - 10u^2 + 22 Now the integrand becomes (u^4 - 10u^2 + 22)(u)(2u du) = 2u^6 - 20u^5 + 44u^2 du That is easy to integrate. Substituting back to x at the end you have 2/21 (3x^2 - 12x + 19) (x+5)^(3/2) + C
*Wednesday, December 14, 2011 at 8:45pm by Steve*

**math**

a. f(t) = e^[tan (pi *t)] Let u = tan (pi t) f(u) = e^u df/dt = df/du du/dt du/dt = pi * sec^2 (pi t) df/du = e^u df/dt = pi* e^[(tan (pi*t)]*sec^2 (pi*t) b) Let u = 8 + ln z f(z) = f[u(z)]) = sqrt u Use the chain rule, as I did in (a) df/dz = df/du*du/dz
*Sunday, October 28, 2007 at 4:37am by drwls*

**Math**

I did x = a tan u dx = a (sec u)^2 du int of a/(a^2 + x^2)^3/2 dx = int of (a sec u)^2/(a^2 + (a tan u)^2)^3/2 du = int of (a sec u)^2/(a sec u)^3 du = int of (cos u)/a du = (sin u)/a + K since u = atan (x/a) = x/(a*(a^2 + x^2)^(1/2)) + K Thanks again...
*Tuesday, January 26, 2010 at 11:23am by Sean*

**Calculus**

looking at the pattern of your answer, it looks like there were 2 terms and the quotient rule was used on the second term looks like you had Q = 19u^2 + (18u+19v(/(u+v)^2 and you took the derivative with respect to u since you know the value of u and v, you can shave down that...
*Monday, October 7, 2013 at 9:05am by Reiny*

**Calculus**

First, I think Reiny dropped a factor of 1/2 in his integral (-x/2, not -x), but that's not a big worry (except that the answer is wrong!) :-) To integrate x√(1-x^2), substitute u=1-x^2 Then you have du = 2x dx x√(1-x^2) = √u du/2 = 1/2 u^(1/2) du integrate ...
*Tuesday, September 11, 2012 at 1:00pm by Steve*

**Calculus**

so I tried again and I got this, don't know if it's good:(I also put the wrong a and b) integral 0 = a and b = pi/3 of sinx dx/(1+cos)^2 u = cosx du= -sinxdx -1 du = sinx dx so: integral -1 du/(1+u)^2 = 1/(u+1) if x = 0, then u = 1 if x = pi/3, then u = 1 so 1/1+1 - 1/1+1 = 0 ...
*Tuesday, August 7, 2012 at 5:09pm by Paul*

**Calculus**

let u=sqrtx du=1/2sqrtx dx or dx=2sqrtx du=2u du yes, inf-constant=inf or divergent
*Thursday, February 28, 2013 at 9:39pm by bobpursley*

**calculus**

let u=2-3z^7, du = -21z^6 dz and you have -1/7 √u du
*Tuesday, September 10, 2013 at 12:16am by Steve*

**Maths**

If you let u = x^3 -6, then du = 2 x^2 dx and the integral can be rewritten as int (1/2) u^3 du which is (1/6)u^4/4 = (1/6)(x^3-6)^4
*Sunday, November 15, 2009 at 10:25am by drwls*

**Calc**

u=x^2+1 du=2x dx integral of 1/2 1/u du=1/2*ln(x^2+1)
*Tuesday, February 15, 2011 at 11:53pm by Johnathon*

**calculus**

f = u^n where u=x^2-6x_23 note that du=2x-6 so f'= nu^(n-1)* du where n= 1.5
*Sunday, April 10, 2011 at 1:58pm by bobpursley*

**Calculus**

u = 1 + cos x du = -sin x dx so we have -du/u
*Tuesday, August 7, 2012 at 5:09pm by Damon*

**calculus**

let u=2x+9 and du = 2dx so you have -csc^2(u) du/2
*Tuesday, September 10, 2013 at 12:17am by Steve*

**calculus**

u=4z-5, du=4 dz and you have cos(u) du/4
*Tuesday, September 10, 2013 at 12:17am by Steve*

**Calc BC**

recall that tan' = sec^2 If we let v = pi/7 x, then dv = pi/7 dx and we have Int[tan^3(v) sec^2(v)] 7/pi dv Now, let u = tan(v). du = sec^2(v) dv Our integral now becomes 7/pi Int[u^3 du] = 7/pi * 1/4 u^4 + C = 7/4pi tan^4(pi/7 x) + C ---------------------- If x = 3tan(u) then...
*Tuesday, September 27, 2011 at 7:24pm by Steve*

**calculus**

du = ? Then substitute u and du into the integral so you are integrating terms of u with respect to u.
*Saturday, January 16, 2010 at 3:07pm by Marth*

**Calculus II**

INT 4^u du what is the integral of a^u du? Ans: a^u / lna check that.
*Tuesday, May 4, 2010 at 3:36pm by bobpursley*

**Integral Help**

u = cosx du = -sinx dx -du/u^-3 1/2cos^2x +c Or sec^2 x /2 + c
*Thursday, November 14, 2013 at 12:44am by Kuai *

**math**

dy/du = (3/2)u^(1/2) du/dx = 4 dy/dx = (dy/du)(du/dx) = (3/2)u^(1/2)(4) = 6u^(1/2) = 6(4x+1)^(/2) = 6√(4x+1)
*Tuesday, June 16, 2009 at 1:35pm by Reiny*

**calculus**

looks very similar to the other one we did. let u = x du/dx = 1 du = dx let dv = sec^2 (9x) dx v = (1/9(tan(9x)) so we have uv - [intg]v du = (x/9)tan(9x) + (1/9)ln(cosx)
*Friday, February 27, 2009 at 10:58pm by Reiny*

**Math Integration**

Integrate: f (x)/(2x + 1) dx let f represent integrate sign let u = x, du = dx => dx = du = f (u)/(2u + 1) du = f u (2u + 1)^(-1) du = (1/2)u^2 (ln|2u + 1|) + c = (1/2)x^2 (ln|2x + 1|) + c ...what did I do wrong? The correct answer is (1/2)x - (1/4)ln|2x + 1| + c
*Wednesday, February 11, 2009 at 4:40pm by Anonymous*

**Calc 1**

note that if u = 1+3sin^2x du = 6sinxcosx so you have ∫[1,4] u^(-1/2) 1/2 du = √u [1,4] = 1
*Friday, November 30, 2012 at 2:18pm by Steve*

**Calculus help please!**

you know that ∫ du/u = ln(u)+C Let u=3x+5 and that's what you have, since du = 3 dx
*Friday, May 3, 2013 at 12:23am by Steve*

**Maths**

Let u=5x+3 du=5dx so you have ∫u^7 (du/5) which should be a piece of cake.
*Tuesday, August 20, 2013 at 2:01pm by Steve*

**math**

g(x) = (8x^2 - 3x)^3 = u^3 use the chain rule: d(g(x))/dx = d(u^3)/du * du/dx =3u^2*du/dx =3(8x^2-3x)²*(16x-3)
*Thursday, March 10, 2011 at 6:38pm by MathMate*

**calculus**

recall that integral secu tanu du = secu Let u=2x and you have du=2dx, so integral 9 secu tanu du = 9secu = 9sec2x
*Tuesday, September 10, 2013 at 12:15am by Steve*

**Calc--derivatives**

Use the function of a function rule, namely dy/dx = dy/dy . du/dx substitute u=log(x) and find the derivative of d(x^u)/du then multiply by du/dx.
*Tuesday, October 13, 2009 at 5:45pm by MathMate*

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