Hints to integrate $\int \frac{1+t^2}{(1-t^2)^2} dt$ Integrate $\int \frac{1+t^2}{(1-t^2)^2} dt \\$
I just want some hint, not the full answer. I've tried substituting $v= 1\pm t^2$ and I've tried  partial fraction but I end up with weird results. Any help? 
 A: Partial fraction decomposition works: $$\int \frac{1+t^2}{(1-t^2)^2} dt  = \int \frac{1+t^2}{(t-1)^2(t+1)^2}\,dt= \int \left(\frac{A}{t+1} + \frac B{(t+1)^2} + \frac C{t-1} + \frac D{(t-1)^2}\right)\,dt$$
$$A(t+1)(t-1)^2 + B(t-1)^2 + C(t-1)(t+1)^2 + D(t+1)^2 = 1+t^2$$
A: First write 
$$\int \frac{1 + t^2}{(1 - t^2)^2} = \int \frac{(1 + t)^2 - 2t}{(1 - t^2)^2}\, dt = \int \frac{(1 + t)^2}{(1 - t^2)^2} + \int \frac{-2t\, dt}{(1 - t^2)^2}.$$
Since $1 - t^2 = (1 - t)(1 + t)$, 
$$\int \frac{(1 + t)^2}{(1 - t^2)^2}\, dt = \int \frac{(1 + t)^2}{(1 - t)^2(1 + t)^2}\, dt = \int \frac{1}{(1 - t)^2}\, dt,$$
which can be evaluated by the $u$-substitution $u = 1 - t$. To evaluate $\int -2t/(1 - t^2)^2\, dt$, use the $v$-substitution $v = 1 - t^2$ and note that $dv = -2t\, dt$.
A: Hint:
$$
\begin{align*}
\int \frac{1+t^2}{(1-t^2)^2}\,{\rm d}t &= 
\int \frac{1-t^2+2t^2}{(1-t^2)^2}\,{\rm d}t\\
&= \int\left[\frac{1-t^2}{(1-t^2)^2}+\frac{2t^2}{(1-t^2)^2}\right]{\rm d}t\\
&= \int\frac{1}{1-t^2}\,{\rm d}t+2\int\frac{t^2}{(1-t^2)^2}\,{\rm d}t.
\end{align*}
$$
Is it simpler now?
A: Maybe use
$\dfrac{1+t^2}{(1-t^2)^2} = \dfrac{1}{2}\dfrac { (1-t)^2 + (1+t)^2}{((1-t)(1+t))^2}$
so you end up integrating $\dfrac{1}{2}\cdot \left( \dfrac{1}{(1+t)^2} + \dfrac{1}{(1-t)^2} \right)$
where each is easy with substitution $1+t = u, 1-t = v$
A: Try the substitution $t = \sin(x)$. This is a challenging integral no matter what you do. If you use this substitution you will have: $$\begin{align}\int\frac{1+\sin^2(x)}{(1-\sin^2(x))^2}\cos(x)dx = \int\frac{\cos(x)+\cos(x)\sin^2(x)}{\cos^4(x)}dx  \\ = \int\frac{1}{\cos^3(x)}+\frac{\sin^2(x)}{\cos^3(x)}dx  \\ = \int\sec^3(x)dx+\int\tan^2(x)\sec(x)dx \\ = \int\sec^3(x)dx+\int(\sec^2(x)-1)\sec(x)dx \\ = \int\sec^3(x)dx+\int\sec^3(x)dx-\int\sec(x)dx \\ = 2\int \sec^3(x) -\int \sec(x)dx \end{align}$$ It is well known (or can be found in an integral table) that $\int\sec(x)dx = \ln(\sec(x)+\tan(x))$. Hence $\int\sec^3(x)dx$ remains to be integrated for a final answer. I recommend integration by parts to calculate this integral, or looking it up in a table.
