does anyone now how to integrate this by partial frac?
$\displaystyle \int\frac{1}{(t^2+1)^3}$
using quickmath wouldnt work
Use the following form: $\displaystyle \frac{A_{1}x + B_1}{ax^2+bx+c} + \frac{A_{2}x + B_2}{(ax^2+bx+c)^2} + \frac{A_{3}x + B_3}{(ax^2+bx+c)^3} $. In this case its $\displaystyle x^2 +1$, $\displaystyle (x^2+1)^2 $, and $\displaystyle (x^2+1)^3 $. Then solve for the coefficients.
So: $\displaystyle \frac{1}{(x^2+1)^3} $ = $\displaystyle \frac{A_{1}x + B_1}{x^2+1} + \frac{A_{2}x + B_2}{(x^2+1)^2} + \frac{A_{3}x + B_3}{(x^2+1)^3} $
$\displaystyle \displaystyle \frac{1}{(t^2+1)^3}$ is already a partial fraction.
You have to integrate by parts.
There is a recurrence relation between $\displaystyle I_n$ and $\displaystyle I_{n-1}$.
$\displaystyle \displaystyle I_n=\int\frac{1}{(x^2+a^2)^n}dx=\frac{1}{a^2}\left[\frac{x}{2(n-1)(x^2+a^2)^{n-1}}+\frac{2n-3}{2(n-1)}I_{n-1}\right],n\geq 2$
$\displaystyle \displaystyle I_n=\frac{1}{a^2}\int\frac{x^2+a^2-x^2}{(x^2+a^2)^n}dx=\frac{1}{a^2}\left[I_{n-1}-\int\frac{x^2}{(x^2+a^2)^n}dx\right]=$
$\displaystyle \displaystyle =\frac{1}{a^2}\left[I_{n-1}+\int x\left(\frac{1}{2(n-1)(x^2+a^2)^{n-1}}\right)'dx\right]$
and you can continue from here.
Use
$\displaystyle \color{blue}\int {\left( {x^2 + a^2 } \right)^m ~dx} = \frac{{x\left( {x^2 + a^2 } \right)^m }}
{{2m + 1}} + \frac{{2a^2 m}}
{{2m + 1}}\int {\left( {x^2 + a^2 } \right)^{m - 1} ~dx},~\forall m\ne-\frac12$
Then if you set $\displaystyle \color{red}m=-n$ will yield red_dog's formula.
This reductions formulas are proved using integration by parts.
Hey Adam:
Perhaps try it this way.
$\displaystyle \int\frac{1}{(x^{2}+1)^{3}}dx$
Let $\displaystyle u=tan^{-1}(x), \;\ x=tan(u), \;\ dx=sec^{2}(u)du$
Make the subs and we get:
$\displaystyle \int\frac{sec^{2}(u)}{(1+tan^{2}(u))^{3}}du$
This equals:
$\displaystyle \int{cos^{4}(u)}du$
Handy reduction formula:
$\displaystyle \int{cos^{n}(u)}du-\frac{cos^{n-1}(u)sin(u)}{n}+\frac{n-1}{n}\int{cos^{n-2}}du$
EDIT: I'm sorry, Adam, I seen you had to use PF. Oh well, there it is anyway.