How do you evaluate the integral of $int (2-x)/(1- x^2)$ ?

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Answer 1 · 2016-01-02T03:43:16

$-1/2 ln|1-x| + 3/2 ln|1+x| +C$ or

$1/2 (3 ln|1+x| -ln|1-x|)+ C$

Use partial fraction decomposition to rewrite the integrant before evaluate the integral

$int(2-x)/(1-x^2) dx = int(2-x)/((1-x)(1+x)) dx$

As partial fraction
$A/(1-x) +B/(1+x) = (2-x)/((1-x)(1+x))$

$A(1+x) + B(1-x) = 2-x$

$x: " " " A -B = -1$
$x^0 " " "A + B = 2$

When you solve the system

$A= 1/2 ; " " " B= 3/2$

Rewrite the integral as

$int1/(2(1-x)) dx + int 3/(2(1+x)) dx$

$-1/2 ln|1-x| + 3/2 ln|1+x| +C$ or

$1/2 (3 ln|1+x| -ln|1-x|)+ C$

How do you evaluate the integral of int (2-x)/(1- x^2)int (2-x)/(1- x^2)?