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I have been trying to evaluate the following integral by using integration by parts, but I contine to yield the incorrect answer.

$$\int \frac{x}{(x+1)^2} dx$$

I choose $u=x$, $dv=\frac{1}{(x+1)^2}dx => du=dx$, $v=\frac{-1}{x+1}$. The integration by parts formula, $\int udv = uv - \int vdu$ yields

$$\int \frac{x}{(x+1)^2} dx = \frac{-x}{x+1} - \int \frac{-1}{x+1}dx = \frac{-x}{x+1}+ ln(x+1)+C$$

However, the integral should be

$$\frac{1}{x+1} + ln(x+1) + C$$

Where did I go wrong? This is my first ask on math stack exchange, so please be kind.

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  • $\begingroup$ Hello, that's one way to do it, but I would like to understand why integration by parts does not work/apply here. $\endgroup$ – DiDoubleTwice Jan 29 '18 at 17:23
  • $\begingroup$ Thanks, all, you guys are fast! $\endgroup$ – DiDoubleTwice Jan 29 '18 at 17:32
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You didn’t do anything wrong. Just notice,

$$\frac{-x}{x+1}=\frac{-x-1+1}{x+1}$$

$$=\frac{1}{x+1}-1$$

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    $\begingroup$ Ah, I see! The -1 is fine as it is a constant absorbed by 'C' as a result of integration. $\endgroup$ – DiDoubleTwice Jan 29 '18 at 17:28
  • $\begingroup$ Yup :) @Didoubletwice $\endgroup$ – Ahmed S. Attaalla Jan 29 '18 at 17:28
  • $\begingroup$ Thanks a plenty! I'll accept your answer as it answers "Where did I go wrong" ;) $\endgroup$ – DiDoubleTwice Jan 29 '18 at 17:31
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Why by parts? $$\frac{x}{(x+1)^2}=\frac{x+1-1}{(x+1)^2}=...$$ and we are done!

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  • $\begingroup$ By parts was the first to come to mind, so I wanted to make sure that I can apply it in a testing scenario. $\endgroup$ – DiDoubleTwice Jan 29 '18 at 17:34
  • $\begingroup$ OK. I understood you. Good luck! $\endgroup$ – Michael Rozenberg Jan 29 '18 at 17:37
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The simplest substitution $u=x+1$ is often overlooked.

$\displaystyle \int \dfrac{x}{(x+1)^2}\mathop{dx}=\int\dfrac{u-1}{u^2}\mathop{du}=\int\left(\dfrac 1u-\dfrac 1{u^2}\right)\mathop{du}=\ln|u|+\dfrac 1u=\ln|x+1|+\dfrac 1{x+1}+C$

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$$g'= (\frac{-x}{x+1}+ ln(x+1)+C)'= \frac {-1(x+1)+x}{(x+1)^2}+\frac 1 {x+1}=\frac x {(x+1)^2}$$

Your answer is correct..

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