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#