How do you integrate int (2x+4)/((x+2)(x-6)(x^2+3)) dx using partial fractions?

1 Answer
May 13, 2018

int(2x+4)/((x+2)(x-6)(x²+3))dx=1/39ln(x-2/13)-1/78ln(x²+3)-2/(13sqrt6)arctan(x/sqrt6)+C,C in RR

Explanation:

int(2x+4)/((x+2)(x-6)(x²+3))dx
=int(2(cancel(x+2)))/((cancel(x+2))(x-6)(x²+3))dx
=2int1/((x-6)(x²+3))dx
Now let 1/((x-6)(x²+3))=A/(x-6)+(Bx+C)/(x²+3)
1/(cancel((x-6)(x²+3)))=(Ax²+3A+Bx²-6Bx+Cx-6C)/(cancel((x-6)(x²+3)))
(A+B)x²+(C-6B)x+3A-6C=1
By identification :
A+B=0[1]
C-6B=0[2]
3A-6C=1[3]

A+B=0[1]
6A+C=0[2]
6A-12C=2[3]

A+B=0[1]
6A+C=0[2]
-13C=2[3]

A+B=0[1]
6A-2/13=0[2]
C=-2/13[3]

B=-1/39[1]
A=1/39[2]
C=-2/13[3]

So:

int(2x+4)/((x+2)(x-6)(x²+3))dx=int(1/(39(x-6))+((-1/39)x-2/13)/(x²+3))dx

=int(1/39(1/(x-2/13))dx-int(1/39x+2/13)/(x²+3)dx

=1/39int1/(x-2/13)dx-1/78int(2x/(x²+3))dx-1/39int(6/(x²+3))dx

=1/39ln(x-2/13)-1/78ln(x²+3)-2/13int(1/(x²+3))dx

=1/39ln(x-2/13)-1/78ln(x²+3)-2/13*1/sqrt6arctan(x/sqrt3)

=1/39ln(x-2/13)-1/78ln(x²+3)-2/(13sqrt6)arctan(x/sqrt6)+C,C in RR

\0/ here's our answer!