1. Partial fractions

does anyone now how to integrate this by partial frac?
$\int\frac{1}{(t^2+1)^3}$

using quickmath wouldnt work

2. Use the following form: $\frac{A_{1}x + B_1}{ax^2+bx+c} + \frac{A_{2}x + B_2}{(ax^2+bx+c)^2} + \frac{A_{3}x + B_3}{(ax^2+bx+c)^3}$. In this case its $x^2 +1$, $(x^2+1)^2$, and $(x^2+1)^3$. Then solve for the coefficients.

So: $\frac{1}{(x^2+1)^3}$ = $\frac{A_{1}x + B_1}{x^2+1} + \frac{A_{2}x + B_2}{(x^2+1)^2} + \frac{A_{3}x + B_3}{(x^2+1)^3}$

3. $\displaystyle \frac{1}{(t^2+1)^3}$ is already a partial fraction.
You have to integrate by parts.
There is a recurrence relation between $I_n$ and $I_{n-1}$.
$\displaystyle I_n=\int\frac{1}{(x^2+a^2)^n}dx=\frac{1}{a^2}\left[\frac{x}{2(n-1)(x^2+a^2)^{n-1}}+\frac{2n-3}{2(n-1)}I_{n-1}\right],n\geq 2$

4. thanks...
hi can you show me how integrating by parts is done?
just the let u's and the let v's

5. $\displaystyle I_n=\frac{1}{a^2}\int\frac{x^2+a^2-x^2}{(x^2+a^2)^n}dx=\frac{1}{a^2}\left[I_{n-1}-\int\frac{x^2}{(x^2+a^2)^n}dx\right]=$
$\displaystyle =\frac{1}{a^2}\left[I_{n-1}+\int x\left(\frac{1}{2(n-1)(x^2+a^2)^{n-1}}\right)'dx\right]$
and you can continue from here.

6. i cant follow the last part

7. Use

$\color{blue}\int {\left( {x^2 + a^2 } \right)^m ~dx} = \frac{{x\left( {x^2 + a^2 } \right)^m }}
{{2m + 1}} + \frac{{2a^2 m}}
{{2m + 1}}\int {\left( {x^2 + a^2 } \right)^{m - 1} ~dx},~\forall m\ne-\frac12$

Then if you set $\color{red}m=-n$ will yield red_dog's formula.

This reductions formulas are proved using integration by parts.

Perhaps try it this way.

$\int\frac{1}{(x^{2}+1)^{3}}dx$

Let $u=tan^{-1}(x), \;\ x=tan(u), \;\ dx=sec^{2}(u)du$

Make the subs and we get:

$\int\frac{sec^{2}(u)}{(1+tan^{2}(u))^{3}}du$

This equals:

$\int{cos^{4}(u)}du$

Handy reduction formula:

$\int{cos^{n}(u)}du-\frac{cos^{n-1}(u)sin(u)}{n}+\frac{n-1}{n}\int{cos^{n-2}}du$

EDIT: I'm sorry, Adam, I seen you had to use PF. Oh well, there it is anyway.