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

Question

1644 views around the world

You can reuse this answer
Creative Commons License

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#