The Negative Integral

Big Idea

Integrating rational functions can get quite challenging unless we split it up into a sum of “smaller” rational functions whose integrals are relatively more trivial to compute.

Questions

Question 1. Evaluate the integral $\displaystyle \int \frac{1}{x^4 - 1}\, \mathrm dx$ .

(Click for Solution)

Solution. Factorise $x^4 - 1 = (x^2 - 1)(x^2 + 1) = (x-1)(x+1)(x^2+1)$ , so that

$\begin{aligned} \frac 1{ (x-1)(x+1)(x^2+1)} &= \frac A{x-1} + \frac B{x+1} + \frac{Cx+D}{x^2+1}. \end{aligned}$

Using the cover-up rule to make the following calculations:

For $x - 1=0$ , set $x = 1$ so that $\displaystyle A = \frac 1{(1+1)(1^2+1)} = 1/4$ .
For $x = -1$ , set $x = -1$ so that $\displaystyle B = \frac 1{ (-1-1)((-1)^2+1)}=-1/4$ .
For $x^2+1 = 0$ , set $x = i$ so that $\displaystyle Ci+D = \frac 1{(i-1)(i+1)} = -1/2$ .

In particular, $C = 0$ and $D = -1/2$ , so that

$\begin{aligned} \int \frac{1}{x^4-1}\, \mathrm dx &= \int \frac 1{ (x-1)(x+1)(x^2+1)}\, \mathrm dx \\ &= \frac 14 \cdot \int \frac 1{x-1}\, \mathrm dx - \frac 14 \cdot \int \frac 1{x+1}\, \mathrm dx - \frac 12 \cdot \int \frac{1}{x^2+1}\, \mathrm dx \\ &= \frac 14 \ln|x-1| - \frac 14 \ln|x+1| - \frac 12 \tan^{-1}(x)+C. \end{aligned}$

Alternate Solution. We observe that

$\begin{aligned} \int \frac{1}{x - 1}\, \mathrm dx &= \int \frac{(x-1)'}{x - 1}\, \mathrm dx =\ln|x-1| + C_1, \\ \int \frac{x}{x^4 - 1}\, \mathrm dx &= \frac 12 \int \frac{(x^2)'}{(x^2)^2 - 1}\, \mathrm dx = \frac 12 \ln \left| \frac{x^2 - 1}{x^2 + 1} \right| + C_2, \\ \int \frac{x^3}{x^4 - 1}\, \mathrm dx &= \frac 14 \int \frac{(x^4-1)'}{x^4-1}\, \mathrm dx = \frac 14 \ln |x^4-1| + C_3. \end{aligned}$

Making the substitution $u = x^2$ , we decompose

$\displaystyle \frac{x^2}{x^4-1} = \frac{u}{(u-1)(u+1)} = \frac A{u-1} + \frac B{u+1}.$

Using the cover-up rule,

$\displaystyle A = \frac{1}{1+1} = \frac 12,\quad B = \frac{-1}{-1-1} = \frac 12.$

Therefore,

$\begin{aligned} \int \frac{x^2}{x^4-1} \mathrm dx &= \frac 12 \int \frac 1{x^2-1} \mathrm dx + \frac 12 \int \frac 1{x^2+1}\, \mathrm dx \\ &= \frac 12 \cdot \frac 12 \ln \left| \frac{x-1}{x+1} \right| + \frac 12 \tan^{-1}(x) + C_4. \end{aligned}$

Finally, we make the clever observation that

$\begin{aligned} (x^4 - 1) &= (x^2 -1)(x^2 + 1) \\ &= (x-1)(x+1)(x^2+1) \\ &= (x-1)(x^3 + x^2 + x + 1), \end{aligned}$

so that

$\begin{aligned} \int \frac 1{x-1}\, \mathrm dx &= \int \frac{x^3 + x^2 + x + 1}{x^4-1}\, \mathrm dx \\ &= \int \frac{x^3}{x^4-1}\, \mathrm dx + \int \frac{x^2}{x^4-1}\, \mathrm dx + \int \frac{x}{x^4-1}\, \mathrm dx + \int \frac{1}{x^4-1}\, \mathrm dx. \end{aligned}$

Therefore,

$\begin{aligned} \int \frac{1}{x^4-1}\, \mathrm dx &= \int \frac 1{x-1}\, \mathrm dx- \int \frac{x^3}{x^4-1}\, \mathrm dx - \int \frac{x^2}{x^4-1}\, \mathrm dx - \int \frac{x}{x^4-1}\, \mathrm dx \\ &= \ln|x-1| - \frac 14 \ln|x^4-1| - \frac 14 \ln \left| \frac{x-1}{x+1} \right| - \frac 12 \tan^{-1}(x) - \frac 12 \ln \left| \frac{x^2 - 1}{x^2 + 1} \right| + C. \end{aligned}$

Question 2. Evaluate the integral $\displaystyle \int \frac{1}{x^4 + 4}\, \mathrm dx$ .

(Click for Solution)

Solution. First make the clever factorisation

$\begin{aligned} x^4 + 4 &= x^4 + 4x^2 + 4 - 4x^2 \\ &= (x^2 + 2)^2 - (2x)^2 \\ &= (x^2 + 2x + 2)(x^2 - 2x +2) \\ &= ((x+1)^2 + 1)((x-1)^2 + 1)\end{aligned}$

Then use the partial fraction decomposition

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

Now the roots of $(x + 1)^2 + 1$ are $-1 \pm i$ . Therefore, using $x = -1 + i$ in the cover-up rule,

$\displaystyle Ai + B = \frac{1}{(-1 + i - 1)^2 + 1} = \frac 18 + \frac 18i \quad \Rightarrow \quad A = B = \frac 18.$

Likewise, the roots of $x^2 - 2x + 2$ are $1 \pm i$ . Therefore, using $x = 1 + i$ in the cover-up rule,

$\displaystyle Ci + D = \frac{1}{(1 + i + 1)^2 + 1} = \frac 18 - \frac 18i \quad \Rightarrow \quad C = -\frac 18,\quad D = \frac 18.$

Therefore,

$\begin{aligned} \int \frac{1}{x^4 + 4}\, \mathrm dx &= \frac 1{16} \int \frac{2(x+1)}{(x+1)^2 + 1} \, \mathrm dx + \frac 18 \int \frac{1}{(x+1)^2 + 1} \, \mathrm dx \\ &\phantom{==} - \frac 1{16} \int \frac{2(x-1)}{(x-1)^2 + 1} \, \mathrm dx + \frac 18 \int \frac{1}{(x-1)^2 + 1}\, \mathrm dx \\ &= \frac 1{16} \ln((x+1)^2 + 1) + \frac 18 \tan^{-1}(x+1) \\ &\phantom{==} - \frac 1{16} \ln((x-1)^2+1) + \frac 18 \tan^{-1}(x-1) + C. \end{aligned}$

—Joel Kindiak, 4 Sept 25, 2239H

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