How do you integrate #int (5x²-2x-1)/( (x+1)(x²+1))# using partial fractions?

Split the fraction up. #x+1# is linear, so its numerator will just be #A#, whereas #x^2+1# is irreducible (over the real numbers) so it will be in the form #Bx+C#.

#(5x^2-2x-1)/((x+1)(x^2+1))=A/(x+1)+(Bx+C)/(x^2+1)#

Which becomes

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

#5x^2-2x-1=Ax^2+A+Bx^2+Bx+Cx+C#

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

Giving the system:

#{(A+B=5),(B+C=-2),(A+C=-1):}#

Subtract the last from the middle to see that #B-A=-1#, which can be added to the first equation to see that #2B=4=>B=2#.

Substitute this to see that #A=3# and #C=-4#.

This gives the partial fraction decomposition of

#(5x^2-2x-1)/((x+1)(x^2+1))=3/(x+1)+(2x-4)/(x^2+1)#

So, we now want to find

#int3/(x-1)dx+int(2x-4)/(x^2+1)dx#

The first integral is simple:

#=3lnabs(x-1)+int(2x-4)/(x^2+1)dx#

In the second, you should realize the #(du)/u# pattern in #(2x)/(x^2+1)#, which is indicative of a natural log. However, this doesn't account for the leftover #-4#.

#=3lnabs(x-1)+int(2x)/(x^2+1)dx-int4/(x^2+1)dx#

We can now deal with the #(du)/u# we identified:

#=3lnabs(x-1)+ln(x^2+1)-int4/(x^2+1)dx#

Recall that the absolute value signs aren't necessary since #x^2+1# is always positive.

To deal with #4/(x^2+1)#, recognize that this is simply #4# times the derivative of the #arctan(x)# function, giving a final answer of

#=3lnabs(x-1)+ln(x^2+1)-4arctan(x)+C#