How do you integrate $x^3/((x^2+5)^2)$ ?

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Answer 1 · 2015-03-26T22:19:04

$int x^3/(x^2+5)^2dx$

We can rewrite :

$x^3/(x^2+5)^2= (betax+gamma)/(x^2+5) + (thetax+nu)/(x^2+5)^2$

So : $((betax+gamma)(x^2+5))/((x^2+5)(x^2+5))+(thetax+nu)/(x^2+5)^2$

Then : $(betax^3+gammax^2+(5beta+theta)x+nu+5gamma)/(x^2+5)^2$

Identification :

$nu + 5gamma = 0$
$5beta + theta = 0$
$gamma = 0$
$beta = 1$

So :

$beta = 1$
$gamma = 0$
$theta = -5$
$nu = 0$

$int x^3/(x^2+5)^2dx = intx/(x^2+5)dx-5int(x)/(x^2+5)^2dx$

$1/2int (2x)/(x^2+5)dx -5/2int(2x)/(x^2+5)^2dx$

$= (1/2ln(x^2+5)+5/(2x^2+10))+C$

How do you integrate x^3/((x^2+5)^2)x^3/((x^2+5)^2)?