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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?

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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:

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

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  • $\begingroup$ I fixed this error, but the answers are still different $\endgroup$ Jan 19 '15 at 1:24
  • $\begingroup$ @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 $+C$. $\endgroup$
    – N. S.
    Jan 19 '15 at 1:28
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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.

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  • $\begingroup$ 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. $\endgroup$ Jan 19 '15 at 1:25

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