Clarification needed: u-sub requires combining 2 constants

[Rear Naked]

Homework Statement

Evaluate ∫-x/(x+1) dx

Homework Equations

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 + CWolfram 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

 

[Mark44]

Homework Statement

Evaluate ∫-x/(x+1) dx

Homework Equations

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.

Two comments:
1. You should connect the expressions you have with =.
2. In these lines --
ln(x+1)-x-1+C
ln(x+1) - x + C
-- C on the 2nd line is different from C on the first. What people usually do is write a different constant (e.g., C') to indicate that the new constant actually is different from the previous one.