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

Start by factoring the denominator.

Now write the partial fraction decomposition.

We now write a system of equations.

Hopefully this helps!

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.