easier ways of solving $\int \frac{4xdx}{(2x+1)^2}$? original integral is
$$\int \frac{4xdx}{(2x+1)^2}$$
I tried partial fractions which worked but it seems to be too long, are there easier ways?
 A: Try splitting as follows and integrate term by term:
$$\int 2\cdot\frac{2x + 1 -1}{(2x+1)^2}$$
Leading to the answer:
$$\ln|2x+1| +(2x+1)^{-1}$$
A: Here is a method which manages to avoid partial fractions by making use of two substitutions.
We begin by letting $x = \frac{1}{2} \tan^2 u, dx = \tan u \cdot \sec^2 u \, du,$ giving
\begin{align*}
\int\frac{4x}{(2x + 1)^2} \, dx &= 2 \int \frac{\tan^3 u \sec^2 u}{(1 + \tan^2 u)^2} \, du\\
&= 2 \int \frac{\tan^3 u}{\sec^2 u} \, du\\
&= 2 \int \frac{\sin^3 u}{\cos u} \, du\\
&= 2 \int \frac{(1 - \cos^2 u)}{\cos u} \sin u \, du.
\end{align*} 
Now let $t = \cos u, dt = -\sin u \, du$. Thus
$$\int \frac{4x}{(2x + 1)^2} \, dx = 2 \int \frac{t^2 - 1}{t} \, dt = t^2 - 2 \ln |t| + C,$$
or
$$\int \frac{4x}{(2x + 1)^2} \, dx = \cos^2 u - 2 \ln |\cos u| + C,$$
since $t = \cos u$. Also, as $\tan^2 u = 2x$ the cosine term in terms of $x$ can be written as $\cos u = 1/\sqrt{2x + 1}$. So finally we have
$$\int \frac{4x}{(2x + 1)^2} = \ln |2x + 1| + \frac{1}{2x + 1} + C,$$
in agreement with the answer given by samjoe.
