7
$\begingroup$

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

I've tried setting $u=e^2$ and $dv=\frac {x}{(1+2x)^2}$ but I'm getting a really long partial answer like:

$$\int \frac {xe^2}{(1+2x)^2}dx = e^{2x}\left[\frac {\ln (1+2x)}{4}+ \frac {1}{4(1+2x)^2}\right] - \ldots$$

I integrated $\frac {x}{(1+2x)^2}$ by decomposing the fraction into $\frac {1}{2(1+2x)}- \frac {1}{2(1+2x)^2}$ which would then give the integral as

$$\int \frac {x}{(1+2x)^2}dx = \int \left[ \frac {1}{2(1+2x)} - \frac {1}{2(1+2x)^2}\right]dx$$

But yes I'm still stuck after doing all of this things. Am I on the right path or should I change the values of $u$ and $dv$?

$\endgroup$
0
1
$\begingroup$

Hint: $~\bigg(\dfrac1t\bigg)'=-\dfrac1{t^2}$

$\endgroup$
1
  • $\begingroup$ That did the job thanks $\endgroup$
    – mopy
    Dec 9 '14 at 22:35

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.