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

Question

Ezra Adams · Accepted Answer

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

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!

Ezekiel Alexander · Answer

undefined To integrate $ \frac{x^2 - 1}{x^3 + x} $ using partial fractions, we first factor the denominator:

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

Since the denominator does not factor further over the real numbers, we can use the following partial fractions decomposition:

$$ \frac{x^2 - 1}{x^3 + x} = \frac{A}{x} + \frac{Bx + C}{x^2 + 1} $$

Multiplying both sides by $ x^3 + x $, we get:

$$ x^2 - 1 = A(x^2 + 1) + (Bx + C)x $$
$$ x^2 - 1 = Ax^2 + A + Bx^2 + Cx $$

Comparing coefficients, we find:

For $ x^2 $: $ A + B = 1 $

For $ x $: $ C = 0 $

For the constant term: $ A = -1 $

Therefore, we have:

$$ \frac{x^2 - 1}{x^3 + x} = \frac{-1}{x} + \frac{1}{x^2 + 1} $$

Now, we can integrate term by term:

$$ \int \frac{x^2 - 1}{x^3 + x} dx = \int \left( \frac{-1}{x} + \frac{1}{x^2 + 1} \right) dx $$
$$ = -\ln|x| + \arctan(x) + C $$

So, the integral of $ \frac{x^2 - 1}{x^3 + x} $ is $ -\ln|x| + \arctan(x) + C $, where $ C $ is the constant of integration.