# Calc.

posted by .

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)

FOIL (u - 1) * sqrt (u)
=((u)^3/2) - ((u)^1/2)
= 1/2 S (((u)^3/2) - ((u)^1/2)) du Now take the derivative and keep the 1/2 on the side.
1/2 ((u)^5/2)/(5/2) - ((u)^3/2)/(3/2) now distribute the 1/2 and bring the 5/2 and 3/2 to the top
1/2 * 2/5 * (u)^5/2 - 1/2 * 2/3 * (u)^ 3/2
cut all the 2 and substitute x^2+1 in the place of u.
1/5 (x^2 - 1)^5/2 - 1/3(x^2 - 1)^3/2

• Calc. -

| x^3*sqrt(x^2 + 1) dx
| = integral sign

| x^2 (sqrt(x^2 + 1)) x dx

u = x^2
du = 2x dx
1/2 du = x dx

1/2 | u (sqrt(u + 1)) du

w = u + 1
dw = du
u = w - 1

1/2 | (w - 1) (sqrt(w)) dw
1/2 | (w - 1) w^1/2 dw
1/2 | w^3/2 dw - 1/2 | w^1/2 dw

1/2 (2/5 w^5/2) - 1/2 ( 2/3 w^3/2) + C
2/10 w^5/2 - 2/6 w^3/2 + C
1/5 w^5/2 - 1/3 w^3/2 + C

w = u + 1
1/5 (u + 1)^5/2 - 1/3 (u + 1)^3/2 + C

u = x^2
1/5 (x^2 + 1)^5/2 - 1/3 (x^2 + 1)^3/2 +
C

## Similar Questions

1. ### calc asap!

can you help me get started on this integral by parts?
2. ### calc check: curve length

Find the length of the curve y=(1/(x^2)) from ( 1, 1 ) to ( 2, 1/4 ) [set up the problem only, don't integrate/evaluate] this is what i did.. let me know asap if i did it right.. y = (1/(x^2)) dy/dx = (-2/(x^3)) L = integral from a …
3. ### calc: avg value

Find the average value of the function "f(x) = x^2 sqrt(1+x^3)" on the interval [0,2]. and this is what i did.. please check for mistakes. thanks :D f(x) = x^2 sqrt(1+x^3), [0,2] f ave = (1/(b-a))*inegral of a to b for: f(x) dx f ave …
4. ### Calc.

Evaluate the definite integral. S b= sqrt(Pi) a= 0 xcos(x^2)dx I'm not sure if this is right?
5. ### calc

find integral using table of integrals ) integral sin^4xdx this the formula i used integral sin^n xdx =-1/n sin^n-1xcosx +n-1/n integral sin^n-2 using the formula this is what i got: integral sin^4xdx=-1/4sin^3xcosx+3/4 integral sin^2xdx= …

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)) …
7. ### Math

1. Evaluate the indefinite integral integral symbol[6x5+2sec(x) tan(x)]dx. 2. Integral symbol 8 at the top, 5 at the bottom 3x2+3x2 dx = Thanks
8. ### Math

1. Evaluate the indefinite integral ([6x^2 + 12x^(3/2) +4x+9]/sqrt x)dx. Answer = + C 2. Evaluate the indefinite integral (12sin x+4tan x)dx. Answer = + C 3. Evaluate the indefinite integral. (x^7)e^(x^8)dx. Answer = + C Thank you …
9. ### Calc BC

1. Find the indefinite integral. Indefinite integral tan^3(pix/7)sec^2(pix/7)dx 2. Find the indefinite integral by making the substitution x=3tan(theta). Indefinite integral x*sqrt(9+x^2)dx 3. Find the indefinite integral. Indefinite …
10. ### Integrals- Log/Ln

Can you check this please i'm not sure why it's wrong .__. 1) indefinite integral of 1/t *ln(t) dt the t*ln(t) is altogether on the bottom i got ln(ln(x)) + C 2) find the indefinite integral of 1/(sqrt(t)*[1-2*sqrt(t)]dt sqrt(x) is …

More Similar Questions