Integrals of Some Particular Functions

7.4 Integrals of Some Particular Functions

This section develops six fundamental integration formulae that serve as building blocks for a wider class of integrals. Each is derived from a carefully chosen substitution, and many complicated-looking integrals can be transformed into one of these standard forms by completing the square or splitting the numerator.

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The Six Standard Formulae

Each of the six results corresponds to a specific algebraic form in the denominator, and each is proved using a substitution that exploits a trigonometric or algebraic identity.

Result (1): $\displaystyle \int \frac{dx}{x^2 - a^2} = \frac{1}{2a} \log \left| \frac{x-a}{x+a} \right| + C$

Derivation

We begin by decomposing the integrand into partial fractions. Notice that:

We want constants $A$ and $B$ such that:

Multiplying both sides by $(x-a)(x+a)$ gives:

To find $A$, set $x = a$: $1 = A(2a)$, so $A = \frac{1}{2a}$.

To find $B$, set $x = -a$: $1 = B(-2a)$, so $B = -\frac{1}{2a}$.

Therefore:

Now integrate term by term:

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Result (2): $\displaystyle \int \frac{dx}{a^2 - x^2} = \frac{1}{2a} \log \left| \frac{a+x}{a-x} \right| + C$

Derivation

This follows the same pattern as Result (1), but note the sign difference in the denominator. Factor $a^2 - x^2 = (a-x)(a+x)$.

We write:

›Proof

To verify the decomposition, set:

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

Multiply through: $1 = A(a+x) + B(a-x)$

Put $x = a$: $1 = A(2a) \implies A = \frac{1}{2a}$

Put $x = -a$: $1 = B(2a) \implies B = \frac{1}{2a}$

Hence the decomposition is correct.

Now integrate:

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Result (3): $\displaystyle \int \frac{dx}{x^2 + a^2} = \frac{1}{a} \tan^{-1}\left(\frac{x}{a}\right) + C$

Derivation

Use the substitution $x = a \tan \theta$. Then $dx = a \sec^2 \theta \, d\theta$.

Since $\tan^2 \theta + 1 = \sec^2 \theta$, this simplifies to:

Now substitute back: $\theta = \tan^{-1}\left(\frac{x}{a}\right)$, giving:

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Result (4): $\displaystyle \int \frac{dx}{\sqrt{x^2 - a^2}} = \log \left| x + \sqrt{x^2 - a^2} \right| + C$

Derivation

Use the substitution $x = a \sec \theta$. Then $dx = a \sec \theta \tan \theta \, d\theta$.

Since $\sec^2 \theta - 1 = \tan^2 \theta$, and for $\theta$ in $(0, \frac{\pi}{2})$ we have $\tan \theta > 0$, so $\sqrt{\tan^2 \theta} = \tan \theta$:

The integral of $\sec \theta$ is a standard result:

Now we need to express $\sec \theta$ and $\tan \theta$ in terms of $x$. From $x = a \sec \theta$, we have $\sec \theta = \frac{x}{a}$. Also:

Therefore:

where $C = C_1 - \log |a|$.

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Result (5): $\displaystyle \int \frac{dx}{\sqrt{a^2 - x^2}} = \sin^{-1}\left(\frac{x}{a}\right) + C$

Derivation

Use the substitution $x = a \sin \theta$. Then $dx = a \cos \theta \, d\theta$.

Since $\sqrt{1 - \sin^2 \theta} = \sqrt{\cos^2 \theta} = |\cos \theta|$, and for $\theta$ in $(-\frac{\pi}{2}, \frac{\pi}{2})$ we have $\cos \theta > 0$, so $|\cos \theta| = \cos \theta$:

Substituting back $\theta = \sin^{-1}\left(\frac{x}{a}\right)$:

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Result (6): $\displaystyle \int \frac{dx}{\sqrt{x^2 + a^2}} = \log \left| x + \sqrt{x^2 + a^2} \right| + C$

Derivation

Use the substitution $x = a \tan \theta$. Then $dx = a \sec^2 \theta \, d\theta$.

Since $\tan^2 \theta + 1 = \sec^2 \theta$, and $\sqrt{\sec^2 \theta} = |\sec \theta|$. For $\theta$ in $(-\frac{\pi}{2}, \frac{\pi}{2})$, $\sec \theta > 0$, so $|\sec \theta| = \sec \theta$:

Now express in terms of $x$. From $x = a \tan \theta$, we have $\tan \theta = \frac{x}{a}$. Also:

Therefore:

where $C = C_1 - \log |a|$.

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Extended Forms: Integrals of Quadratic Expressions

The six standard results above are for denominators of the form $x^2 \pm a^2$ or $\sqrt{x^2 \pm a^2}$. But many integrals involve quadratic expressions $ax^2 + bx + c$. The strategy is to complete the square and then use a substitution to reduce the integral to one of the six standard forms.

Method (7): $\displaystyle \int \frac{dx}{ax^2 + bx + c}$

Procedure

1. Write $ax^2 + bx + c = a\left[x^2 + \frac{b}{a}x + \frac{c}{a}\right]$
2. Complete the square: $= a\left[\left(x + \frac{b}{2a}\right)^2 + \left(\frac{c}{a} - \frac{b^2}{4a^2}\right)\right]$
3. Let $t = x + \frac{b}{2a}$, so $dx = dt$
4. Let $k^2 = \frac{c}{a} - \frac{b^2}{4a^2}$ (or $-k^2$ if the expression is negative)

The integral then becomes:

which matches either Result (1), (2), or (3) depending on the sign.

Method (8): $\displaystyle \int \frac{dx}{\sqrt{ax^2 + bx + c}}$

Procedure …