# Hard Polinomic Integral

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

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

thx

• 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. – 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

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