Calculating $\int \frac{dx}{(x^2+1)^3}$ integral Would anyone help me calculate the following integral? $\int \frac{dx}{(x^2+1)^3}$
During our lecutre we've done very similiar one, $\int \frac{dx}{(x^2+1)^2}$ like that: 
$\int \frac{dx}{(x^2+1)^2} = \int \frac{x^2+1-x^2}{(x^2+1)^2}dx = \int \frac{1}{x^2+1}dx -  \int \frac{x^2}{(x^2+1)^2}dx = $
$= \Biggr\rvert \begin{equation}
\begin{split}
& u = x \quad v' =\frac{x}{(x^2+1)^2} =\frac{1(x^2+1)'}{2(x^2+1)^2}\\
& u' = 1 \quad v = -\frac{1}{2} \frac{1}{x^2+1}
\end{split}
\end{equation} \Biggr\rvert$
$= \arctan x - (-x\frac{1}{2}\frac{1}{x^2+1} + \frac{1}{2} \int \frac{dx}{x^2+1})$
$= \arctan x + \frac{x}{2(x^2+1)} - \frac{1}{2}\arctan x + C = \frac{1}{2}\arctan x + \frac{x}{2(x^2+1)} + C$
Thank you.
 A: So why not use the same trick? Since you know from lecture what $$f_2(x) = \int \frac{dx}{\left(x^2+1\right)^2}$$ is, you now have
$$
\begin{split}
f_3(x) 
 &= \int \frac{dx}{\left(x^2+1\right)^3}\\
 &= \int \frac{dx}{\left(x^2+1\right)^2} - \int x \cdot \frac{xdx}{\left(x^2+1\right)^3}\\
 &= f_2(x) - \text{proceed by parts as before}
\end{split}
$$
A: We will find a general reduction formula for the integral
$$I_n=\int\frac{dx}{(ax^2+b)^n}$$
Integration by parts with 
$$dv=dx\Rightarrow v=x\\ u=\frac1{(ax^2+b)^n}\Rightarrow du=\frac{-2anx}{(ax^2+b)^{n+1}}dx$$
Yields 
$$I_n=\frac{x}{(ax^2+b)^n}+2n\int\frac{ax^2}{(ax^2+b)^{n+1}}dx$$
$$I_n=\frac{x}{(ax^2+b)^n}+2n\int\frac{ax^2+b}{(ax^2+b)^{n+1}}dx-2bn\int\frac{dx}{(ax^2+b)^{n+1}}$$
$$I_n=\frac{x}{(ax^2+b)^n}+2n\int\frac{dx}{(ax^2+b)^{n}}-2bnI_{n+1}$$
$$I_n=\frac{x}{(ax^2+b)^n}+2nI_n-2bnI_{n+1}$$
$$2bnI_{n+1}=\frac{x}{(ax^2+b)^n}+(2n-1)I_n$$
$$I_{n+1}=\frac{x}{2bn(ax^2+b)^n}+\frac{2n-1}{2bn}I_n$$
replacing $n+1$ with $n$,
$$I_{n}=\frac{x}{2b(n-1)(ax^2+b)^{n-1}}+\frac{2n-3}{2b(n-1)}I_{n-1}$$
Now for the base case $I_1$:
$$I_1=\int\frac{dx}{ax^2+b}$$
Let $$x=\sqrt{\frac{b}a}\tan u\Rightarrow dx=\sqrt{\frac{b}a}\sec^2u\, du$$
So $$I_1=\sqrt{\frac{b}a}\int\frac{\sec^2u}{b\tan^2u+b}du$$
$$I_1=\frac1{\sqrt{ab}}\int\frac{\sec^2u}{\sec^2u}du$$
$$I_1=\frac1{\sqrt{ab}}\int du$$
$$I_1=\frac{u}{\sqrt{ab}}$$
$$I_1=\frac1{\sqrt{ab}}\arctan\sqrt{\frac{a}{b}}x+C$$
Plug in your specific $a,b$ and $n$, and you're good to go.

Edit: Whatever I'll just give you the answer.
Plugging in $a=1,\ b=1,\ n=3$ we have 
$$I_{3}=\frac{x}{4(x^2+1)^{2}}+\frac{3}{4}I_{2}$$
Note that 
$$I_{2}=\frac{x}{2(x^2+1)}+\frac{1}{2}I_{1}$$
$$I_{2}=\frac{x}{2(x^2+1)}+\frac{1}{2}\arctan x+C$$
So 
$$I_{3}=\frac{x}{4(x^2+1)^{2}}+\frac{3x}{8(x^2+1)}+\frac{3}{8}\arctan x+C$$
A: Use $$\int \frac{dx}{(x^2+1)^3}=\frac{x}{4(x^2+1)^2}+\frac{3}{4}\int \frac{dx}{(x^2+1)^2}$$ and then use the value for the integral you already calculated.
A: I used:
$$p=x^2+1\to dx =\frac{1}{2\sqrt{p-1}}$$
which leads to:
$$\int{\frac{dx}{(x^2+1)^3}=\frac12 \int{\frac{dp}{p\sqrt{p-1}}}}$$
Then the substitution $$q=\sqrt{p-1}\to dp=2q\space dq$$
Leads to $$\frac12\int\frac{dp}{p\sqrt{p-1}}=\frac12\int\frac{2q}{q(q^2+1)}dq=\int\frac{dq}{q^2+1}$$
Which is resolved simply.
