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

Question

Christopher Allers · Accepted Answer

# -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#

Isaiah Allen · Answer

undefined To evaluate the integral of $\int \frac{2-x}{1-x^2} \, dx$, you can use partial fraction decomposition. First, rewrite the integrand as $\frac{A}{1+x} + \frac{B}{1-x}$, where $A$ and $B$ are constants to be determined. Then, integrate each term separately, and solve for $A$ and $B$ by equating coefficients. Finally, integrate the resulting expression. The integral of $\frac{2-x}{1-x^2}$ is equal to $\ln|1+x| - \ln|1-x| + C$, where $C$ is the constant of integration.

Ezekiel Alexander · Answer

undefined To evaluate the integral of $ \int \frac{2-x}{1-x^2} $, you use partial fraction decomposition.