How do you use partial fractions to find the integral #int (x^2-1)/(x^3+x)dx#?

1 Answer
Feb 28, 2017

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

Explanation:

Start by factoring the denominator.

#x^3 + x = x(x^2 + 1)#

Now write the partial fraction decomposition.

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

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

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

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

We now write a system of equations.

#{(A + C = 1), (B = 0), (C = -1):}#

Solving, we get #A = 2, B = 0, C = -1#.

#int(2x)/(x^2 + 1) - 1/xdx#

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

We now make the substitution #u = x^2 + 1#. Then #du = 2xdx# and #dx = (du)/(2x)#.

#int (2x)/u * (du)/(2x) - int 1/xdx#

#int 1/u du - int 1/x dx#

#ln|u| - ln|x| + C#

#ln|x^2 + 1| - ln|x| + C#

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

Hopefully this helps!