How do you solve this integral?

Question

#intdx/(x^2-1)^2#

1166 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-03-28T11:29:33

#int\ ("d"x)/(x^2-1)^2#
#=1/4(ln(x+1)-ln(x-1)-(2x)/(x^2-1))+C#

#int\ ("d"x)/(x^2-1)^2#
#=int\ ("d"x)/((x+1)^2(x-1)^2)#

Now, let's do the partial fractions. Assume that

#1/((x+1)^2(x-1)^2)=A/(x+1)+B/(x+1)^2+C/(x-1)+D/(x-1)^2#

for some constants #A,B,C,D#.

Then,
#1=A(x+1)(x-1)^2+B(x-1)^2+C(x+1)^2(x-1)+D(x+1)^2#

Expand to get
#1=(A+C)x^3+(B+C+D-A)x^2+(2D-2B-A-C)x+A+B-C+D#.

Equate coefficients:
#{(A+C=0),(B+C+D-A=0),(2D-2B-A-C=0),(A+B-C+D=1):}#

Solving gives #A=B=D=1/4# and #C=-1/4#.

Thus, our original integral is
#int\ (1/(4(x+1))+1/(4(x+1)^2)-1/(4(x-1))+1/(4(x-1)^2))\ "d"x#
#=1/4ln(x+1)-1/(4(x+1))-1/4ln(x-1)-1/(4(x-1))+C#

Simplify:
#=1/4(ln(x+1)-ln(x-1)-(2x)/(x^2-1))+C#