How do you evaluate the indefinite integral of #dx/(81+x^2)^2#?

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Answer 1 · 2015-05-11T10:06:29

Hi, here the polynome is irreducible you can't do partial fraction so we will use substitution

Let's #x=9tan(u)#
#dx = 9sec^2(u) du#
#u=arctan(1/9x)#

So we have :

#int1/(81+x^2)^2dx = int1/(81+81tan^2(u))^2*sec^2(u)du#

#=1/7291int1/(1+tan^2(u))^2*sec^2(u)du#

Don't forget #1+tan^2(u) = sec^2(u)#

#1/729intcos^4(u)*sec^2(u)du = 1/729intcos^2(u)du#

Don't forget #cos^2(u) = 1/2(1+cos(2u))#

#1/1458int1+cos(2u)du = 1/2916int2+2cos(2u)du#

#1/2916[2u+sin(2u)]+C#

#1/2916[2arctan(1/9x)+sin(2arctan(1/9x))]#

#sin(arctan(x)) = x/sqrt(x^2+1)#

so we have

#1/1458[arctan(1/9x)+(9x)/sqrt(x^2+81)]+C#