How do you integrate #int (x^2-1)/((x)*(x^2+1))# using partial fractions?

1 Answer
Narad T.
Oct 18, 2016

#=ln((x^2+1)/x)+C#

Explanation:

#(x^2-1)/(x(x^2+1))=A/x+(Bx+C)/(x^2+1)#
Developing
#(x^2-1)/(x(x^2+1))=(A(x^2+1)+ x(Bx+C))/(x(x^2+1)#
So we can compare the coefficients
#x^2-1=A(x^2+1)+ x(Bx+C)#
For #x^2# , #1=A+B# and #-1=A# and #0=C#
Hence #A=-1#and #B=2#
#int((x^2-1)dx)/(x(x^2+1))=intAdx/x+int((Bx+C)dx)/(x^2+1)#
#=int(-1dx)/x+int((2x)dx)/(x^2+1#
#=-lnx +ln(x^2+1)+C#
#=ln((x^2+1)/x)+C#

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