Integration by parts - hint I'm stuck on a passage on my textbook:
$$ \int \frac{1}{(1+t^2)^3} dt = \frac{t}{4(t^2+1)^2}+\frac{3}{4} \int \frac{1}{(t^2+1)^2} dt$$
I know that it should be easy but I just can't figure out what is the product of functions considered in this integration by parts..
can you help me?
thanks a lot
 A: Notice, in general, use integration by parts as follows $$\int\frac{1}{(1+x^2)^{n-1}}\ dx=\int \underbrace{\frac{1}{(1+x^2)^{n-1}}}_{I}\cdot \underbrace{1}_{II}\ dx$$
$$=\frac{1}{(1+x^2)^{n-1}}\int 1\ dx-\int\left(\frac{d}{dx}\frac{1}{(1+x^2)^{n-1}}\cdot \int 1\ dx\right)\ dx$$
$$=\frac{x}{(1+x^2)^{n-1}}-\int\left(-\frac{2(n-1)x^2}{(1+x^2)^{n}}\right)\ dx$$
$$=\frac{x}{(1+x^2)^{n-1}}+2(n-1)\int\left(\frac{1+x^2-1}{(1+x^2)^{n}}\right)\ dx$$
$$=\frac{x}{(1+x^2)^{n-1}}+2(n-1)\int \frac{1}{(1+x^2)^{n-1}}\ dx-2(n-1)\int \frac{1}{(1+x^2)^{n}}\ dx$$
$$\int \frac{1}{(1+x^2)^{n}}\ dx=\frac{x}{2(n-1)(1+x^2)^{n-1}}+\frac{2n-3}{2(n-1)}\int\frac{1}{(1+x^2)^{n-1}}\ dx$$
Above reduction formula which can be easily applied to find 
$\int\frac{1}{(1+x^2)^{3}}\ dx$ as follows 
$$\int\frac{1}{(1+x^2)^{3}}\ dx=\frac{x}{4(1+x^2)^{2}}+\frac 34\int \frac{1}{(1+x^2)^{2}}\ dx$$
$$=\frac{x}{4(1+x^2)^{2}}+\frac 34\left(\frac{x}{2(1+x^2)}+\frac 12\int \frac{1}{1+x^2}\ dx\right)$$
$$=\frac{x}{4(1+x^2)^{2}}+\frac{3x}{8(1+x^2)}+\frac{3}{8}\tan^{-1}(x)+C$$
A: Alternatively you can make use of Lebnitz Integral formula since you already know  $\int \frac{1}{x^2+1}\, dx= \arctan x$
Consider $$I(x) = \int \frac{dt}{x^2+t^2}=\frac{\arctan x}{x}+C$$
Note that $$I''(x)=8 x^2\int \frac{dt}{(x^2+t^2)^3}$$
Guess you can figure out the rest
A: $$
\int \frac{1}{(1+t^2)^3} \, dt = \overbrace{\int u\,dt = tu - \int t\,du}^\text{integration by parts}. 
$$
$$
u = \frac 1 {(1+t^2)^3}, \qquad du = \Big(\cdots\cdots{?}\cdots\cdots\Big)\, dt
$$
Fill in the blank and go on from there.
A: $$\int \dfrac {1}{(t^2+1)^3}\,dt = \int \dfrac {(1+t^2) - t^2}{(t^2+1)^3}\, dt = \int \dfrac {1}{(t^2+1)^2}\,dt + \dfrac{1}{4}\int t\, d(t^2+1)^{-2}$$$$ = \int \dfrac {1}{(t^2+1)^2}\,dt + \dfrac{1}{4}\dfrac {t}{(t^2+1)^2} - \dfrac{1}{4}\int  \dfrac{1}{(t^2+1)^2}\,dt $$$$ = \dfrac{1}{4}\dfrac {t}{(t^2+1)^2} + \dfrac{3}{4}\int  \dfrac{1}{(t^2+1)^2}\,dt  $$
