Evaluate $ \int_0^3 \frac{x^3}{1-x^4}\, dx. $ Evaluate
$$
\int_0^3 \frac{x^3}{1-x^4}\, dx.
$$ 
I evaluated the integral and got $\left[\dfrac{-\ln(1-x^4)}{4}\right]_0^3$ which ended up diverging. Any help is appreciated!
 A: The integral gives up easily to direct attack, but with a little twist. 
Use the obvious substitution $u=x^4$:
$$\int_0^3 \frac{x^3}{1-x^4}\, dx=\frac{1}{4} \int_0^{81} \frac{du}{1-u}=$$
Now it makes sense to separate the integral into two parts, so the denominator is positive for each part:
$$=\frac{1}{4} \int_0^1 \frac{du}{1-u}-\frac{1}{4} \int_1^{81} \frac{du}{u-1}=$$
Now in the first integral we use the substitution $t=1-u$ and in the second $p=u-1$:
$$=\frac{1}{4} \int_0^1 \frac{dt}{t}-\frac{1}{4} \int_0^{80} \frac{dp}{p}=\frac{1}{4} \int_0^1 \frac{dt}{t}-\frac{1}{4} \int_0^1 \frac{dp}{p}-\frac{1}{4} \int_1^{80} \frac{dp}{p}=$$
But the first two divergent integrals just cancel each other out (they are equal), so we are allowed to write:
$$\int_0^3 \frac{x^3}{1-x^4}\, dx=-\frac{1}{4} \int_1^{80} \frac{dp}{p}=-\frac{1}{4} \ln 80=-\frac{1}{4} (\ln 5+4 \ln 2)$$
This is the correct answer (formally). Although Wolfram Alpha writes that the original integral diverges, and the answer we obtained is just Cauchy principal value of the integral.

Edit. A little justification for the calcelling the divergent integrals. The problem is at $t \to 0$, thus:
$$\int_0^1 \frac{dt}{t}=\lim_{\epsilon \to 0} \int_\epsilon^1 \frac{dt}{t}$$
In this case the limit doesn't exist.
But for any positive real number $\epsilon>0$ we have formally:
$$\int_\epsilon^1 \frac{dt}{t} \equiv \int_\epsilon^1 \frac{dp}{p}$$
We just renamed the variable after all. Moreover, in our particular example we have $t=-p$. So setting:
$$\lim_{\epsilon \to 0} \int_\epsilon^1 \frac{dt}{t}-\lim_{\epsilon \to 0} \int_\epsilon^1 \frac{dp}{p}=\lim_{\epsilon \to 0} \left( \int_\epsilon^1 \frac{dt}{t}-\int_\epsilon^1 \frac{dp}{p} \right)=0$$
Makes perfect sense.
But the rigorous justification should involve contour integration in the complex plane.
