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#