How can you integrate this #int ((3x+5)dx) /((x-1)^2(x+1))#?

#int ((3x+5)dx) /((x-1)^2(x+1))#

I need steps after this step #A/(x-1)+B/(x-1)^2 + C/(x+1)#

1 Answer
Harish Chandra Rajpoot
Jul 26, 2018

#1/2\ln|{x+1}/{x-1}|-4/{x-1}+C#

Explanation:

Let

#\frac{3x+5}{(x-1)^2(x+1)}=\frac{A}{x-1}+\frac{B}{(x-1)^2}+\frac{C}{x+1}#

#3x+5=A(x^2-1)+B(x+1)+C(x-1)^2#

#3x+5=(A+C)x^2+(B-2C)x-A+B+C#

Comparing the corresponding coefficients on both the sides , we get

#A+C=0 \ ............(1)#

#B-2C=3 \ ............(2)#

#-A+B+C=5 \ ............(3)#

Adding (1) & (3), we get

#A+C-A+B+C=0+5#

#B+2C=5\ ...........(4)#

Adding (2) & (4) we get

#B-2C+B+2C=3+5#

#2B=8#

#B=4#

setting #B=4# in (2), we get

#4-2C=3#

#C=1/2#

Setting #C=1/2# in (1), we get

#A+1/2=0#

#A=-1/2#

Now, the partial fractions are as follows

#\frac{3x+5}{(x-1)^2(x+1)}=\frac{-1/2}{x-1}+\frac{4}{(x-1)^2}+\frac{1/2}{x+1}#

Integrating above equation w.r.t. #x# as follows

#\int \frac{3x+5}{(x-1)^2(x+1)}\ dx#

#=\int \frac{-1/2}{x-1}\ dx+\int \frac{4}{(x-1)^2}\ dx+\int \frac{1/2}{x+1}\ dx#

#=-1/2\int \frac{1}{x-1}\ dx+4\int (x-1)^{-2}\ dx+1/2\int \frac{1}{x+1}\ dx#

#=-1/2\ln|x-1|+4(-1/(x-1))+1/2\ln|x+1|+C#

#=1/2\ln|{x+1}/{x-1}|-4/{x-1}+C#

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