# An easy integral

1. May 10, 2010

### jasper10

An "easy" integral

1. The problem statement, all variables and given/known data

solve ∫x/(x+1)

3. The attempt at a solution

let u = x + 1 ,therefore, x = u - 1

hence, ∫x/(x+1) = ∫(u-1)/u = ∫u/u - ∫1/u = ∫1 - ∫1/u = u - lnu = x + 1 - ln(x+1)
=x+1+ln(1/(x+1))

differentiating this gives me x + 2 which does not make sense!

2. May 10, 2010

### Cyosis

Re: An "easy" integral

Use the chain rule.

3. May 10, 2010

### Dick

Re: An "easy" integral

Differentiating that does NOT give you x+2. Try differentiating again. Carefully, this time.

4. May 10, 2010

### Staff: Mentor

Re: An "easy" integral

Why not leave it as x + 1 - ln(x + 1)? I don't see any advantage in rewriting the last term as ln(1/(x + 1)), and it makes taking the derivative more complicated.

Also, don't forget the constant of integration!

5. May 10, 2010

### jasper10

Re: An "easy" integral

So you agree that x + 1 - ln(x+1) = x+1+ln(1/(x+1)) ?

what would be the steps for differentiating x+1+ln(1/(x+1)) without changing it back to : x + 1 - ln(x+1) ?

Thanks!

6. May 10, 2010

### jasper10

Re: An "easy" integral

Oh ok thanks, doesn't matter, i got it!

(At night I can think clearer than in the daytime!)