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

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

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