How do you find $int (x+2)/((x^2+2)x) dx$ using partial fractions?

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Answer 1 · 2017-05-02T14:05:36

$int(1/x-(x-1)/(x^2+2))dx=lnx-1/2ln|x^2+2|+1/sqrt2tan^(-1)(x/sqrt2)+c$

Partial fractions of $(x+2)/((x^2+2)x)$ would be of the form

$(x+2)/((x^2+2)x)=A/x+(Bx+C)/(x^2+2)$

I.e. $x+2=A(x^2+2)+x(Bx+C)$

Comparing coefficients of $x*2$ , $x$ and constant term

we have $A+B=0$ , $C=1$ and $2A=2$

i.e. $A=1$ , $B=-1$ and $C=1$

i.e. $(x+2)/((x^2+2)x)=1/x-(x-1)/(x^2+2)$

Hence our integral becomes $int(1/x-(x-1)/(x^2+2))dx$

= $int1/xdx-intx/(x^2+2)dx+int1/(x^2+2)dx$

= $lnx-1/2ln|x^2+2|+1/sqrt2tan^(-1)(x/sqrt2)+c$

How do you find int (x+2)/((x^2+2)x) dxint (x+2)/((x^2+2)x) dx using partial fractions?