Integrating $\frac{x^3}{(81-x^2)^2}$ I've been trying to figure out this integral for an hour or so now, but keep failing.  I can't figure out where I go wrong:
$$I = \int \frac{x^3}{(81-x^2)^2} dx$$
Let $x = 9sin\theta \implies dx = 9 \cos \theta d \theta$
$$I =\int \frac{9^3 \sin^3 \theta}{9^4(1-\sin^2 \theta)^2} 9 \cos \theta d\theta$$
$$I = \int \tan^3 \theta d\theta$$
$$I = \int \tan \theta \sec^2 \theta d \theta - \int \tan \theta d\theta$$
$$I = \frac{1}{2}tan^2 \theta + \ln| \cos \theta | + C$$
Using $cos \theta = \frac{\sqrt{81-x^2}}{9}$, $\tan \theta = \frac{x}{\sqrt{81-x^2}}$
$$I = \frac{1}{2}(\frac{x^2}{81-x^2}) + \frac{1}{2} \ln |\frac{81-x^2}{81}|+C$$
Somewhere I went terribly wrong, but I just can't figure out where.  I've done this problem around 5 times now.

Alternatively write $81-x^2=y$
$$\dfrac{x^3}{(81-x^2)^2}=\dfrac{81-(81-x^2)}{(81-x^2)^2}\cdot x $$
