# Thread: Integral problem

1. ## Integral problem

So I did this integral problem (attachment), but I didn't get the same answer as the book. Looks like the book made the substitution u= x^(1/2) +1, but I don't see how that makes the problem any easier... and I don't see were I did anything wrong.

Any help? Thanks!

2. ## Re: Integral problem

I tried to follow your working but I'm afraid I got lost. In these situations, it's best to just forget it and start again.

I'm assuming you want to stick with your substitution so I'm going to try it.

$\displaystyle \int{\frac{x}{\sqrt{x}+1}} dx$

Letting $\displaystyle u=\sqrt{x}$ gives $\displaystyle du=\frac{1}{2\sqrt{x}} dx \Rightarrow dx=2\sqrt{x}~du \Rightarrow dx=2u~du$

The integral becomes:

$\displaystyle \int\frac{u^2}{u+1}\cdot 2u~du$

$\displaystyle =2\int\frac{u^3}{u+1}~du$

$\displaystyle =2\int\frac{u^3+1-1}{u+1}~du$

$\displaystyle =2\int[\frac{u^3+1}{u+1}-\frac{1}{u+1}]~du$

$\displaystyle =2\int[\frac{(u+1)(u^2-u+1)}{u+1}-\frac{1}{u+1}]~du$ using the sum of two cubes formula.

$\displaystyle =2\int{(u^2-u+1}-\frac{1}{u+1})~du$

Which you can now integrate term by term.

Edit: I notice, upon reflection, that you already had this exact thing. See my following post.

3. ## Re: Integral problem

Whoops, I didn't notice that this is exactly what you had. As far as I can tell, your answer is fine. What did the book have?