1
$\begingroup$

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?

$\endgroup$
  • 2
    $\begingroup$ Try setting $$2x+1=y$$ $\endgroup$ – lab bhattacharjee Oct 29 '17 at 11:51
  • $\begingroup$ @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. $\endgroup$ – Junaid Aftab Oct 29 '17 at 12:24
1
$\begingroup$

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

Leading to the answer:

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

$\endgroup$
  • 1
    $\begingroup$ (...which is partial fractions.) $\endgroup$ – Hans Lundmark Oct 29 '17 at 12:14
  • $\begingroup$ @Hans He said partial fractions seems too long, but I wanted to say that it isn't that long after all :) $\endgroup$ – samjoe Oct 30 '17 at 8:58
  • $\begingroup$ I can't argue with that! :-) $\endgroup$ – Hans Lundmark Oct 30 '17 at 13:42
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.