Integrating x/(x+1) with a Simple Substitution

[jasper10]

An "easy" integral

Homework Statement

solve ∫x/(x+1)

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!
Please help!

 

[Cyosis]

Use the chain rule.

 

[Dick]

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

 

[Mark44]

Homework Statement

solve ∫x/(x+1)

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))

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!

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

 

[jasper10]

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!

 

[jasper10]

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

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