Chapter 7 · Integrals

Example 14: $\int\frac{x^2}{(x^2+1)(x^2+4)}\,dx$

ExamplesPartial fractions · Section 7.5

The question

Find

\int \frac{x^2}{(x^2+1)(x^2+4)}\,dx

Exam answer

Let $I = \displaystyle\int\frac{x^2\,dx}{(x^2 + 1)(x^2 + 4)}$ ⁠.

Put $x^2 = y$ in the integrand and let

\frac{y}{(y + 1)(y + 4)} = \frac{A}{y + 1} + \frac{B}{y + 4}

\Rightarrow\ y = A(y + 4) + B(y + 1)

Putting $y = -1$ ⁠: $-1 = 3A$ ⁠, so $A = -\frac13$ ⁠. Putting $y = -4$ ⁠: $-4 = -3B$ ⁠, so $B = \frac43$ ⁠.

Using $\displaystyle\int\frac{dx}{x^2 + a^2} = \frac1a\tan^{-1}\frac xa + C$ ⁠,

\begin{aligned} \therefore\ I &= -\frac13\int\frac{dx}{x^2 + 1} \\ &\quad + \frac43\int\frac{dx}{x^2 + 2^2} \\ &= -\frac13\tan^{-1}x \\ &\quad + \frac43\cdot\frac12\tan^{-1}\frac x2 + C \end{aligned}

$\therefore\ I = -\dfrac13\tan^{-1}x + \dfrac23\tan^{-1}\dfrac x2$ $+ C$ ⁠.

Want the reason behind every step, the basics and the checks?

The one idea this problem is built on

The integrand contains $x$ only through $x^2$ ⁠. So for the algebra of splitting the fraction we may call $x^2$ a single letter $y$ ⁠: then $\frac{y}{(y+1)(y+4)}$ has two different linear factors and splits easily. Afterwards we put $x^2$ back, and each piece $\frac{k}{x^2 + a^2}$ integrates to an inverse tangent. The letter $y$ is only a shorthand for the algebra; we never integrate with respect to $y$ ⁠.

Given, to find, and the plan

Given: $\frac{x^2}{(x^2+1)(x^2+4)}$ ⁠. Both factors of the bottom are at least 1, so the integrand is defined for every real $x$ ⁠.
To find: Its indefinite integral.
Plan: (1) Put $x^2 = y$ in the fraction. (2) Split $\frac{y}{(y+1)(y+4)}$ ⁠. (3) Put $y = x^2$ back. (4) Integrate with the formula for $\int\frac{dx}{x^2+a^2}$ ⁠. (5) Differentiate back.

Basics you need

1Partial fractions for two different linear factors

Table 7.2, row 1: $\frac{py+q}{(y-a)(y-b)} = \frac{A}{y-a} + \frac{B}{y-b}$ ⁠, $a \ne b$ ⁠. Basics: splitting fractions

2The inverse tangent integral

Section 7.4, formula (3): $\int \frac{dx}{x^2+a^2} = \frac1a\tan^{-1}\frac{x}{a} + C$ (for $a > 0$ ⁠). With $a = 1$ this is standard formula (x), $\int\frac{dx}{1+x^2} = \tan^{-1}x + C$ ⁠. Lesson: integrals of particular functions

3Derivative of an inverse tangent

$\frac{d}{dx}\tan^{-1}u = \frac{1}{1+u^2}\cdot\frac{du}{dx}$ ⁠, used in the final check. Basics: the chain rule backwards

Solution

Write the fraction in terms of $y = x^2$

Put $x^2 = y$ ⁠. Then
$\frac{x^2}{(x^2+1)(x^2+4)} = \frac{y}{(y+1)(y+4)}.$
In terms of $x$ the bottom has quadratic factors, which would need the form $\frac{Ax+B}{x^2+1} + \frac{Cx+D}{x^2+4}$ with four unknowns. Because only even powers of $x$ appear, the letter $y$ turns the same fraction into one with two linear factors and just two unknowns. This is purely algebra: the identity we find holds for every value of $y$ (other than $-1$ and $-4$ ⁠), in particular for every $y = x^2 \ge 0$ ⁠.

Replacing a repeated expression by a single letter is a relabelling, not a change of the integration variable. Basics: splitting fractions
Split into partial fractions

The fraction is proper in $y$ (top degree 1, bottom degree 2). By Table 7.2, row 1,
$\frac{y}{(y+1)(y+4)} = \frac{A}{y+1} + \frac{B}{y+4},$ $y = A(y+4) + B(y+1).$
Comparing coefficients of $y$ and the constants: $A + B = 1$ and $4A + B = 0$ ⁠. Subtracting, $3A = -1$ ⁠, so
$A = -\frac13, \qquad B = \frac43.$
Cross-check: $y = -1$ gives $-1 = 3A$ ⁠; $y = -4$ gives $-4 = -3B$ ⁠. The same values.

Two equations, two unknowns. Substituting $y = -1$ and $y = -4$ into the polynomial identity confirms both values independently.

An identity between polynomials holds for every value, so its coefficients match. Lesson: partial fractions
Put $x^2$ back
$\frac{x^2}{(x^2+1)(x^2+4)} = -\frac{1}{3(x^2+1)} + \frac{4}{3(x^2+4)}.$
The identity in $y$ holds for every $y \ge 0$ ⁠, so replacing $y$ by $x^2$ gives an identity valid for every real $x$ ⁠. The integration is now done with respect to $x$ ⁠, as in the question.

Quick test at $x = 0$ ⁠: left side $0$ ⁠, right side $-\frac13 + \frac13 = 0$ ⁠. Basics: splitting fractions
Integrate with the inverse tangent formula
$\begin{aligned} \int \frac{x^2\,dx}{(x^2+1)(x^2+4)} &= -\frac13\int\frac{dx}{x^2+1} \\ &\quad + \frac43\int\frac{dx}{x^2+2^2}. \end{aligned}$
By formula (3) of Section 7.4 with $a = 1$ and with $a = 2$ ⁠:
$\int\frac{dx}{x^2+1} = \tan^{-1}x, \qquad \int\frac{dx}{x^2+4} = \frac12\tan^{-1}\frac{x}{2}.$
So
$\begin{aligned} &\int \frac{x^2\,dx}{(x^2+1)(x^2+4)} \\ &= -\frac13\tan^{-1}x + \frac43\cdot\frac12\tan^{-1}\frac x2 + C \\ &= -\frac13\tan^{-1}x + \frac23\tan^{-1}\frac x2 + C. \end{aligned}$
The pieces are $\frac{1}{x^2 + a^2}$ ⁠, not $\frac{1}{x - a}$ ⁠, so they give inverse tangents, not logarithms. Write $4$ as $2^2$ to read off $a = 2$ ⁠; forgetting the factor $\frac1a = \frac12$ is the usual slip.

Formula (3): $\int\frac{dx}{x^2+a^2} = \frac1a\tan^{-1}\frac xa + C$ ⁠. Lesson: integrals of particular functions
Check: differentiate the answer

By the chain rule, $\frac{d}{dx}\tan^{-1}\frac x2 = \frac{1}{1 + x^2/4}\cdot\frac12 = \frac{2}{x^2+4}$ ⁠. Hence
$\begin{aligned} &\frac{d}{dx}\left[-\frac13\tan^{-1}x + \frac23\tan^{-1}\frac x2\right] \\ &= -\frac{1}{3(x^2+1)} + \frac{4}{3(x^2+4)} \\ &= \frac{-(x^2+4) + 4(x^2+1)}{3(x^2+1)(x^2+4)} \\ &= \frac{3x^2}{3(x^2+1)(x^2+4)} = \frac{x^2}{(x^2+1)(x^2+4)}. \end{aligned}$
The derivative returns exactly the integrand, so the answer is correct for every real $x$ ⁠.

$\frac{d}{dx}\tan^{-1}x = \frac{1}{1+x^2}$ (formula (x) read backwards) and the chain rule. Basics: integration undoes differentiation

Result

\begin{aligned} &\int \frac{x^2}{(x^2+1)(x^2+4)}\,dx \\ &= -\frac13\tan^{-1}x + \frac23\tan^{-1}\frac x2 + C, \end{aligned}

valid for every real $x$ ⁠. This is the book's answer.

Check it with numbers

At a point. At $x = 1$ ⁠: the integrand is $\frac{1}{2\cdot 5} = 0.1$ ⁠; the derivative of our answer is $-\frac16 + \frac{4}{15} = \frac{1}{10}$ ⁠.

Over an interval. On $[0, 2]$ ⁠:

\begin{aligned} F(2) - F(0) &= -\frac13\tan^{-1}2 + \frac23\cdot\frac\pi4 \\ &= 0.1545492030\ldots \end{aligned}

Numerical integration of the integrand over $[0, 2]$ gives $0.1545492030\ldots$ ⁠, the same.

Common mistakes

Integrating with respect to $y$ ⁠. The letter $y$ was only for the algebra; $dx$ is not $dy$ (in fact $dy = 2x\,dx$ ⁠).
Writing logarithms. $\int\frac{dx}{x^2+1}$ is $\tan^{-1}x$ ⁠, not $\log|x^2+1|$ ⁠.
Losing the $\frac1a$ ⁠. $\int\frac{dx}{x^2+4} = \frac12\tan^{-1}\frac x2$ ⁠, not $\tan^{-1}\frac x2$ ⁠.

What to take away

If only $x^2$ appears, call it $y$ to find the partial fractions with less work, then put $x^2$ back.
$\frac{k}{x^2+a^2}$ integrates to $\frac ka\tan^{-1}\frac xa$ ⁠.
The book's next example (Example 15) is different: there the substitution is used for the integration itself.

Every fact used, and where it comes from

Fact	Used in	Source
Partial fractions, two different linear factors	Step 2	Table 7.2, row 1, Section 7.5 (Lesson: partial fractions)
Equal polynomials have equal coefficients	Step 2	Algebra (Class XI)
Property V (term by term)	Step 4	Section 7.2.1
$\int\frac{dx}{x^2+a^2} = \frac1a\tan^{-1}\frac xa + C$	Step 4	Formula (3), Section 7.4 (Lesson: integrals of particular functions)
$\frac{d}{dx}\tan^{-1}x = \frac{1}{1+x^2}$ ⁠; chain rule	Step 5	Standard formula (x), Section 7.2; Chapter 5

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Example 14: ∫x2(x2+1)(x2+4) dx\int\frac{x^2}{(x^2+1)(x^2+4)}\,dx∫(x2+1)(x2+4)x2​dx

Basics you need

Solution

Write the fraction in terms of y=x2y = x^2y=x2

Split into partial fractions

Put x2x^2x2 back

Integrate with the inverse tangent formula

Check: differentiate the answer

Every fact used, and where it comes from

Example 14: $\int\frac{x^2}{(x^2+1)(x^2+4)}\,dx$

Write the fraction in terms of $y = x^2$

Put $x^2$ back