Different answer for integral for two different methods

Question

I am trying to integrate $\frac{1-x}{(x+1)^2}$, but I get to answers for two different methods:

First, $\frac{1-x}{(x+1)^2} = \frac{1-x+1-1}{(x+1)^2} = \frac{2}{(x+1)^2} - \frac{x+1}{(x+1)^2} = \frac{2}{(x+1)^2} - \frac{1}{(x+1)}$. Integrating this easily gives $\frac{-2}{(x+1)} - \ln|x+1|$.

Second method: Integration by parts. $u = (1-x)$, $dv = \frac{1}{(1+x)^2}$. $uv - \int v du$ is then equal to $\frac{x-1}{(x+1)} - \ln|x+1|$.

How do I resolve this contradictory conclusions?

score 0 · Answer 1 · answered Jan 19 '15 at 01:21

0

One of your answers is wrong. The way to verify an integration is to differentiate the alleged antiderivative, and see that it is the integrand.

answered Jan 19 '15 at 01:21

Robert Israel

470,583

Yes, and I am aware that the first answer is the correct one. But I am not sure what is wrong in the second method. – Sarayevo Clemens Jan 19 '15 at 01:25

score 0 · Accepted Answer · answered Jan 19 '15 at 01:23

0

You made a mistake in the second method:

For the integration by parts you calculated $udv -\int v du $ instead of $uv -\int v du $.

Added: After fixing this error the answers are the same:

$$\frac{x-1}{(x+1)} - \ln|x+1|+ C=1-\frac{2}{(x+1)}- \ln|x+1| +C=-\frac{2}{(x+1)}- \ln|x+1| +C$$

answered Jan 19 '15 at 01:23

N. S.

134,609

I fixed this error, but the answers are still different – Sarayevo Clemens Jan 19 '15 at 01:24
@SarayevoClemens They are the same, you forgot the $+C$. Note that $\frac{x-1}{x+1}=\frac{x+1-2}{x+1}=1-\frac{2}{x+1}$ and $1$ goes inside $+C$. – N. S. Jan 19 '15 at 01:28

Different answer for integral for two different methods

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