Integration by Partial Fractions (Exercise 7.5, Part 2)

This lesson continues Exercise 7.5, integrating rational functions by partial fractions. Several of these integrals have a numerator whose degree equals the denominator's, so the first step is polynomial division, and then the proper remainder is split into partial fractions. The lesson also shows one integral solved without partial fractions, and a general shortcut for a special family of integrals.

Question 6: divide first

We want

$\int \frac{1 - x^2}{x(1 - 2x)}\, dx.$

The numerator and denominator both have degree two, so we divide first. Writing the denominator as $-2x^2 + x$ and dividing gives quotient $\tfrac{1}{2}$ and a proper remainder:

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

So

$I = \int \frac{1}{2}\, dx + \int \frac{\tfrac{1}{2}x - 1}{x(1 - 2x)}\, dx = \frac{1}{2}x + I_2.$

Partial fractions for the remainder

Let

$\frac{\tfrac{1}{2}x - 1}{x(1 - 2x)} = \frac{A}{x} + \frac{B}{1 - 2x},$

so $\tfrac{1}{2}x - 1 = A(1 - 2x) + Bx$.

At $x = 0$: $-1 = A$, so $A = -1$.

At $x = \tfrac{1}{2}$: $\tfrac{1}{4} - 1 = B \cdot \tfrac{1}{2}$, so $-\tfrac{3}{4} = \tfrac{1}{2}B$ and $B = -\tfrac{3}{2}$.

Therefore

$I_2 = \int \frac{-1}{x}\, dx + \int \frac{-\tfrac{3}{2}}{1 - 2x}\, dx = -\log|x| + \frac{3}{4}\log|1 - 2x|,$

where the $\tfrac{3}{4}$ comes from $-\tfrac{3}{2}$ divided by the $-2$ from the derivative of $1 - 2x$.

Putting it together,

$I = \frac{1}{2}x - \log|x| + \frac{3}{4}\log|1 - 2x| + C.$

Question 7: another same-degree case

We want

$\int \frac{x^2 + x + 1}{x^2 - 1}\, dx.$

Again the degrees match, so divide. Writing the denominator as $x^2 + 0x - 1$ and dividing gives quotient $1$ and remainder $x + 2$:

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

So $I = x + I_2$, where

$I_2 = \int \frac{x + 2}{(x + 1)(x - 1)}\, dx.$

Partial fractions

Let

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

so $x + 2 = A(x - 1) + B(x + 1)$.

At $x = 1$: $3 = 2B$, so $B = \tfrac{3}{2}$.

At $x = -1$: $1 = -2A$, so $A = -\tfrac{1}{2}$.

Therefore

$I_2 = -\frac{1}{2}\log|x + 1| + \frac{3}{2}\log|x - 1|,$

and

$I = x - \frac{1}{2}\log|x + 1| + \frac{3}{2}\log|x - 1| + C.$

The same integral without partial fractions

The remainder integral can also be done by splitting it:

$I_2 = \int \frac{x}{x^2 - 1}\, dx + \int \frac{2}{x^2 - 1}\, dx.$

First piece. Put $x^2 - 1 = t$, so $2x\, dx = dt$ and $x\, dx = \tfrac{1}{2}\, dt$:

$\int \frac{x}{x^2 - 1}\, dx = \frac{1}{2}\int \frac{dt}{t} = \frac{1}{2}\log|x^2 - 1|.$

Second piece. Using $\displaystyle\int \frac{1}{x^2 - a^2}\, dx = \frac{1}{2a}\log\left|\frac{x - a}{x + a}\right|$ with $a = 1$:

$\int \frac{2}{x^2 - 1}\, dx = \log\left|\frac{x - 1}{x + 1}\right|.$

The two forms of the answer agree.

A general shortcut

Consider

$\int \frac{1}{x(x^n + 1)}\, dx.$

Multiply the top and bottom by $x^{n-1}$:

$I = \int \frac{x^{n-1}}{x^n(x^n + 1)}\, dx.$

Put $x^n = t$, so $n x^{n-1}\, dx = dt$ and $x^{n-1}\, dx = \tfrac{1}{n}\, dt$:

$I = \frac{1}{n}\int \frac{dt}{t(t + 1)}.$

By partial fractions $\dfrac{1}{t(t + 1)} = \dfrac{1}{t} - \dfrac{1}{t + 1}$, so

$I = \frac{1}{n}\big(\log|t| - \log|t + 1|\big) = \frac{1}{n}\log\left|\frac{t}{t + 1}\right|.$

Resubstituting $t = x^n$,

$I = \frac{1}{n}\log\left|\frac{x^n}{x^n + 1}\right| + C.$

Worked case with $x^4 - 1$

For

$\int \frac{1}{x(x^4 - 1)}\, dx,$

multiply top and bottom by $x^3$:

$I = \int \frac{x^3}{x^4(x^4 - 1)}\, dx.$

Put $x^4 = t$, so $4x^3\, dx = dt$ and $x^3\, dx = \tfrac{1}{4}\, dt$:

$I = \frac{1}{4}\int \frac{dt}{t(t - 1)}.$

Here $\dfrac{1}{t(t - 1)} = -\dfrac{1}{t} + \dfrac{1}{t - 1}$, so

$I = \frac{1}{4}\big(-\log|t| + \log|t - 1|\big) = \frac{1}{4}\log\left|\frac{t - 1}{t}\right|.$

Resubstituting $t = x^4$,

$I = \frac{1}{4}\log\left|\frac{x^4 - 1}{x^4}\right| + C.$

Key takeaways

• When the numerator and denominator have the same degree, divide first, then apply partial fractions to the proper remainder.
• Find partial fraction constants quickly by substituting the values of $x$ that zero out each factor.
• For $\displaystyle\int \frac{1}{x(x^n \pm 1)}\, dx$, multiply top and bottom by $x^{n-1}$ and substitute $t = x^n$ to reduce it to a simple partial fraction in $t$.