Integration by Parts - $\int \frac{x}{(x+1)^2} dx$

Question

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.

Answer 1

You didn’t do anything wrong. Just notice,

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

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

Answer 2

Why by parts? $$\frac{x}{(x+1)^2}=\frac{x+1-1}{(x+1)^2}=...$$ and we are done!

Answer 3

The simplest substitution $u=x+1$ is often overlooked.

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

Answer 4

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

Your answer is correct..