Question #64159

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Answer 1 · 2017-04-17T13:48:05

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

Expand the square:

$=int(x+1/x)(x+1/x)dx$

$=int(x^2+x(1/x)+x(1/x)+(1/x)^2)dx$

$=int(x^2+2+x^-2)dx$

Integrate these using $intx^ndx=x^(n+1)/(n+1)$ and $aintdx=ax$ :

$=x^3/3+2x+x^-1/(-1)+C$

$=x^3/3+2x-1/x+C$

Getting a common denominator:

$=(x^4+6x^2-3)/(3x)+C$