How do you use partial fractions to find the integral $int (x^2+x+2)/(x^2+2)^2dx$ ?

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Answer 1 · 2018-03-04T09:21:24

The answer is $=1/sqrt(2)arctan(x/sqrt2)-1/(2(x^2+2))+C$

We need

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

Perform the decomposition into partial fractions

$(x^2+x+2)/(x^2+2)^2=(x^2+2+x)/(x^2+2)^2$

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

Therefore,

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

The first integral is

$int(1dx)/(x^2+2)=int(dx)/(2((x/sqrt2)^2+1))$

$=1/2*sqrt2arctan(x/sqrt2)$

$=1/sqrt(2)arctan(x/sqrt2)$

For the second integral

Let $u=x^2+2$ , $=>$ , $du=2xdx$

Therefore,

$int(xdx)/(x^2+2)^2=1/2int(du)/(u^2)$

$=-1/2*1/u$

$=-1/2*1/(x^2+2)$

Finally,

$int((x^2+x+2)dx)/(x^2+2)^2=1/sqrt(2)arctan(x/sqrt2)-1/(2(x^2+2))+C$

How do you use partial fractions to find the integral int (x^2+x+2)/(x^2+2)^2dxint (x^2+x+2)/(x^2+2)^2dx?