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#