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#