# 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?

• Try setting $$2x+1=y$$ – lab bhattacharjee Oct 29 '17 at 11:51
• @labbhattacharjee has a nice suggestion. Let me try and elaborate it. The substitution would allow you to expresses the numerator as a linear combination of polynomials of at most degree one (since $x$ is of degree one). The computational advantage, however, is given by the observation that denominator now gets converted to a polynomial of degree two, which is not a linear combination of polynomials. Now you can split up the fraction term by term and integrate. – Junaid Aftab Oct 29 '17 at 12:24

Try splitting as follows and integrate term by term: $$\int 2\cdot\frac{2x + 1 -1}{(2x+1)^2}$$

$$\ln|2x+1| +(2x+1)^{-1}$$
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.