# Calculus III

posted by on .

I'm supposed to find the expression 'y', when

(x^2+100)y'=23

When y(10sqrt(3))=0

Also, I have to find the indefite integral
of

xsqrt(19x-7), using integration tables in the back of the book "Calculus of a Single Variable, 8th Edition." Tables are in appendix B

I used the formula

[2/[b(2n-3)]]*([u^n]*(a+bu)^[3/2]-na[integral of [u^(n-1)sqrt[a+bu]]]

I've used
a=-7
b=19
u=x
du=xd
But it doesn't seem to work.

• Calculus III - ,

Unless you really have to finish a problem in a hurry, never use integration tables or other tools for problem solving. Only use them to very the solution when you are done.

(x^2+100)y'=23 ---->

dy/dx = 23/[x^2 + 100] ----->
y = Integral of dy =

Integral of 23 dx/[x^2 + 100], looks like an arctan to me.

Integral of x sqrt(19x-7)dx ?

Write the integral in terms of functins you do know the inegral of. Rewrite the factor of x as follows:

x = 1/19 (19 x) =

1/19 (19 x - 7 + 7) =

1/19 (19 x - 7) + 7/19

I think you'll now see the answer. These sorts of tricks cannot be learned by just looking in integral tables.

• Calculus III - ,

I got the first one, thanks for that.

But on the second one, are you saying that x=(1/19)*(19x-7)+(7/19)?

If so, would I just multiply the entire thing out, or what?