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

1 Answer
Nov 6, 2015

I= -1/4ln(x^2+3)+1/(2sqrt3)arctan(x/sqrt3) +1/2ln|x+3| +C

Explanation:

(3-x)/((x^2+3)(x+3))=(Ax+B)/(x^2+3)+C/(x+3)=

=((Ax+B)(x+3)+C(x^2+3))/((x^2+3)(x+3))=

=(Ax^2+Bx+3Ax+3B+Cx^2+3C)/((x^2+3)(x+3))=

=((A+C)x^2+(B+3A)x+(3B+3C))/((x^2+3)(x+3))

A+C=0 => A=-C
B+3A=-1 => B=-1+3C
3B+3C=3 => B+C=1

-1+3C+C=1 => 4C=2 => C=1/2

A=-1/2

B=1-C=1/2

I=int (3-x)/((x^2+3)(x+3))dx = int (-1/2x+1/2)/(x^2+3)dx + 1/2intdx/(x+3)

I=-1/4int(2xdx)/(x^2+3)+1/2intdx/(x^2+(sqrt3)^2)+1/2ln|x+3|

I= -1/4ln(x^2+3)+1/(2sqrt3)arctan(x/sqrt3) +1/2ln|x+3| +C