# Clarification needed: u-sub requires combining 2 constants

## Homework Statement

Evaluate ∫-x/(x+1) dx

## The Attempt at a Solution

I do a U-sub

u=x+1
du=dx
x=u-1

-∫(u-1)/u du

-∫[u/u - 1/u] du

-∫1 - 1/u du

-[∫1du - ∫1\u du]

-[u - ln(u) + C]

ln(u)-u+C

ln(x+1)-(x+1)+C

ln(x+1)-x-1+C

I combine -1 and C into a single C to get the final answer

ln(x+1) - x + C

Wolfram says the answer is correct, but its method is slightly different. It uses long division at the outset to go from -∫x/(x+1) dx → -∫1 - 1/(x+1) dx before seperating into 2 integrals, and doing a u-sub on the second.

My answer ends up the same, but getting that extra -1 from substituting the (x+1) in for u is making me nervous for some reason, even though I ultimately combine it with the constant. Is my process equally correct (I think it is) or did I just get lucky with the solution?

Thanks

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## Homework Statement

Evaluate ∫-x/(x+1) dx

## The Attempt at a Solution

I do a U-sub

u=x+1
du=dx
x=u-1

-∫(u-1)/u du

-∫[u/u - 1/u] du

-∫1 - 1/u du

-[∫1du - ∫1\u du]

-[u - ln(u) + C]

ln(u)-u+C

ln(x+1)-(x+1)+C

ln(x+1)-x-1+C

I combine -1 and C into a single C to get the final answer

ln(x+1) - x + C

Wolfram says the answer is correct, but its method is slightly different. It uses long division at the outset to go from -∫x/(x+1) dx → -∫1 - 1/(x+1) dx before seperating into 2 integrals, and doing a u-sub on the second.

My answer ends up the same, but getting that extra -1 from substituting the (x+1) in for u is making me nervous for some reason, even though I ultimately combine it with the constant. Is my process equally correct (I think it is) or did I just get lucky with the solution?

Thanks
Either approach can be taken to produce the antiderivative. Many times it is the case that more than one technique can be used.