To break the fraction
{2-x}/{(1-x)(1+x^2)}2−x(1−x)(1+x2)
into partial fractions, we assume
{2-x}/{(1-x)(1+x^2)} = A/{1-x} + {Bx +C}/{1+x^2}2−x(1−x)(1+x2)=A1−x+Bx+C1+x2
Multiplying both sides by (1-x)(1+x^2)(1−x)(1+x2) yields the equation
2-x = A(1+x^2) +Bx(1-x)+C(1-x)2−x=A(1+x2)+Bx(1−x)+C(1−x)
= (A+C) +(B-C) x +(A-B)x^2=(A+C)+(B−C)x+(A−B)x2
Comparing the coefficients of powers of xx on both sides gives
A+C = 2A+C=2, B-C=-1, B−C=−1, and A-B=0A−B=0.
It is easy to see that the solution of this set of equations is A=B=1/2, C=3/2A=B=12,C=32, so that
{2-x}/{(1-x)(1+x^2)} = 1/{2(1-x)} + {1/2x +3/2}/{1+x^2}2−x(1−x)(1+x2)=12(1−x)+12x+321+x2
So the required integral is
int {2-x}/{(1-x)(1+x^2)} dx ∫2−x(1−x)(1+x2)dx
= int {dx}/{2(1-x)} + 3/2 int {dx}/{1+x^2} + 1/4 int {2xdx}/{1+x^2} =∫dx2(1−x)+32∫dx1+x2+14∫2xdx1+x2
= -1/2 ln |1-x|+1/4 ln(1+x^2) +3/2 tan^{-1} x+c=−12ln|1−x|+14ln(1+x2)+32tan−1x+c
