Help with Integral Homework: Reduce to Solve

[ArmandStarks]

Homework Statement

The problem is this:
∫[√x /(1+x)] dx
I used sustitution method
u= √x
u^2=x
2udu=dx

Homework Equations

My new integral is:
∫[(u*2udu)/(1+u^2)]du
2∫[u^2/(1+u^2)]du
I need help to reduce this point to continue, I guess I need some algebraic steps

The Attempt at a Solution

The book shows: x^3/3 -3x - 10Ln|x-5| + C as the answer but I think this is wrong because I don't have x-5 anywhere

 

[Mark44]

Homework Statement

The problem is this:
∫[√x /(1+x)] dx
I used sustitution method
u= √x
u^2=x
2udu=dx

Homework Equations

My new integral is:
∫[(u*2udu)/(1+u^2)]du
2∫[u^2/(1+u^2)]du

This looks OK. I would use polynomial long division to turn this improper rational expression into a nicer form for integration. If you're not familiar with this technique, do a web search for "polynomial long division".

I need help to reduce this point to continue, I guess I need some algebraic steps

The Attempt at a Solution

The book shows: x^3/3 -3x - 10Ln|x-5| + C as the answer but I think this is wrong because I don't have x-5 anywhere

I agree - the book's answer doesn't look right.

 

[SammyS]

Homework Statement

The problem is this:
∫[√x /(1+x)] dx
I used sustitution method
u= √x
u^2=x
2udu=dx

Homework Equations

The Attempt at a Solution

My new integral is:
∫[(u*2udu)/(1+u^2)]du
2∫[u^2/(1+u^2)]du
I need help to reduce this point to continue, I guess I need some algebraic steps

The book shows: x^3/3 -3x - 10Ln|x-5| + C as the answer but I think this is wrong because I don't have x-5 anywhere

Hello ArmandStarks. Welcome to PF !

You can use "long division" to divide u² by (u² + 1) .

Or use the following "trick" .

$\displaystyle \frac{u^2}{1+u^2}=\frac{u^2+1-1}{1+u^2}$

$\displaystyle =\frac{1+u^2}{1+u^2}-\frac{1}{1+u^2}$​

 

[ArmandStarks]

Hello ArmandStarks. Welcome to PF !

You can use "long division" to divide u² by (u² + 1) .

Or use the following "trick" .

$\displaystyle \frac{u^2}{1+u^2}=\frac{u^2+1-1}{1+u^2}$

$\displaystyle =\frac{1+u^2}{1+u^2}-\frac{1}{1+u^2}$​

I just found that "trick" and the 2 integrals are easy to do.
Thank you both!