How do you find #int (x+1)/(x(x^2-1)) dx# using partial fractions?

Question

Emma Aldridge · Accepted Answer

You try to split the rational function into a sum that will be really easy to integrate.

First of all : #x^2 - 1 = (x-1)(x+1)#. Partial fraction decomposition allows you to do that : #(x+1)/(x(x^2 - 1)) = (x+1)/(x(x-1)(x+1)) = 1/(x(x-1)) = a/x + b/(x-1)# with #a,b in RR# that you have to find. In order to find them, you have to multiply both sides by one of the polynomials at the left of the equality. I show one example to you, the other coefficient is to be found the same way. We're gonna find #a# : we have to multiply everything by #x# in order to make the other coefficient disappear. #1/(x(x-1)) = a/x + b/(x-1) iff 1/(x-1)= a + (bx)/(x-1)#. #x = 0 iff -1 = a# You do the same thing in order to find #b# (you multiply everything by #(x-1)# then you choose #x = 1#), and you find out that #b = 1#. So #(x+1)/(x(x^2 - 1)) = 1/(x-1) - 1/x#, which implies that #int(x+1)/(x(x^2 - 1))dx = int(1/(x-1) - 1/x)dx = intdx/(x-1) - intdx/x = lnabs(x-1) - lnabsx#

Emilia Aguilar · Answer

undefined To find the integral of $\frac{x+1}{x(x^2-1)}$ using partial fractions, we first decompose the fraction into partial fractions:

$$
\frac{x+1}{x(x^2-1)} = \frac{A}{x} + \frac{B}{x+1} + \frac{C}{x-1}
$$

Next, we clear the denominators by multiplying both sides by $x(x^2-1)$:

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

Expanding and collecting like terms:

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

Grouping terms with the same powers of $x$:

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

This equation yields the following system of equations:

$$
\begin{align*}
A + B + C &= 0 \
-B + C &= 1 \
-A &= 1
\end{align*}
$$

Solving this system, we find $A = -1$, $B = 0$, and $C = 1$.

Substitute these values back into the partial fraction decomposition:

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

Now, integrate each term separately:

$$
\int \frac{-1}{x} dx + \int \frac{1}{x-1} dx
$$

This gives:

$$
-\ln|x| + \ln|x-1| + C
$$

Where $C$ is the constant of integration.