Hi guys can someone help me with this integral... I try diferents substitution but can't solve it. Need a hint pls!

$$ \int \frac{x}{e^{3x}(1-x)^4}dx\,. $$


  • 2
    $\begingroup$ Write the integral as $\int e^{-3x} (1-x)^{-4} \mathrm{d}x - \int e^{-3x}(1-x)^{-3}\mathrm{d}x$, and integrate the first of these by parts. The solution conveniently pops out. $\endgroup$ – stochasticboy321 Sep 1 '16 at 2:14

You can do this with two substitutions as follows:

$$\int \frac{x}{e^{3x}(1-x)^4} dx = \int \frac{\ln u}{u^4 (1-\ln u)^4} du$$

After making the substitution $e^x = u$. Then further rewriting:

$$\int \frac{\ln u}{(u(1-\ln u))^4} du$$

We are now set up to make the second substitution. Write $u(1-\ln u) = v.$ Then from this it follows that $\frac{dv}{du}= - \ln u$. So now we have:

$$\int\frac{-1}{v^4} dv = \frac{1}{3v^3} + C$$.

Chasing backwards through the definitions, we have $$\frac{1}{3(u(1-\ln u))^3} +C = \frac{1}{3(e^x(1-x))^3} +C$$


If you're satisfied with a brute force approach. Let $u = 1-x$.

$$\int \frac{x}{e^{3x}(1-x)^4} \> dx \implies \int \frac{u-1}{e^{3(1-u)}u^4} \> du$$

This is just

$$\int (e^{3u-3}u^{-3}-e^{3u-3}u^{-4}) \> du$$

We notice we get lucky if we integrate by parts and differentiate $u^{-3}$. In particular we get

$$\frac{1}{3} e^{3u-3}u^{-3} - \int (\frac{1}{3}e^{3u-3}(-3)u^{-4}) \> du \> \> - \int e^{3u-3}u^{-4} \> du$$

We notice the last two terms cancel out and our solution is just

$$\frac{1}{3(e^x(1-x))^3} + C$$


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.