How do you find $int (2-x)/((1-x)(1+x^2)) dx$ using partial fractions?

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Answer 1 · 2016-07-16T07:22:55

$-1/2 ln |1-x|+1/4 ln(1+x^2) +3/2 tan^{-1} x+c$

To break the fraction
${2-x}/{(1-x)(1+x^2)}$
into partial fractions, we assume

${2-x}/{(1-x)(1+x^2)} = A/{1-x} + {Bx +C}/{1+x^2}$

Multiplying both sides by $(1-x)(1+x^2)$ yields the equation

$2-x = A(1+x^2) +Bx(1-x)+C(1-x)$
$= (A+C) +(B-C) x +(A-B)x^2$

Comparing the coefficients of powers of $x$ on both sides gives

$A+C = 2$ , $B-C=-1,$ and $A-B=0$ .

It is easy to see that the solution of this set of equations is $A=B=1/2, C=3/2$ , so that

${2-x}/{(1-x)(1+x^2)} = 1/{2(1-x)} + {1/2x +3/2}/{1+x^2}$

So the required integral is

$int {2-x}/{(1-x)(1+x^2)} dx$
$= int {dx}/{2(1-x)} + 3/2 int {dx}/{1+x^2} + 1/4 int {2xdx}/{1+x^2}$
$= -1/2 ln |1-x|+1/4 ln(1+x^2) +3/2 tan^{-1} x+c$

How do you find int (2-x)/((1-x)(1+x^2)) dxint (2-x)/((1-x)(1+x^2)) dx using partial fractions?