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

Answer 1

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

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 turns into

#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#

Providing the framework with:

#{(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 provides the partial fraction breakdown for

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

Thus, we are now looking for

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

The initial integral is straightforward:

#=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#
Sign up to view the whole answer

By signing up, you agree to our Terms of Service and Privacy Policy

Sign up with email
Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

Not the question you need?

Drag image here or click to upload

Or press Ctrl + V to paste
Answer Background
HIX Tutor
Solve ANY homework problem with a smart AI
  • 98% accuracy study help
  • Covers math, physics, chemistry, biology, and more
  • Step-by-step, in-depth guides
  • Readily available 24/7