Integration of Rational Algebraic Functions

Rational algebraic integrands built from quadratic expressions $ax^2+bx+c$ (or their square roots) do not yield to the substitution and by-parts methods developed earlier in the chapter directly -- they need their own toolkit, organised here into four types that build on one another, one feeding into the next.

Type I. These are the six "base case" integrals built from $a^2\pm x^2$ and $x^2\pm a^2$, and every later type is eventually reduced to one of these six:

The first two are proved by factoring the denominator as a difference of squares and splitting into partial fractions: $\dfrac1{a^2-x^2}=\dfrac1{(a-x)(a+x)}=\dfrac1{2a}\left[\dfrac1{a+x}+\dfrac1{a-x}\right]$, so integrating term by term gives $\dfrac1{2a}\big[\log|a+x|-\log|a-x|\big]+c=\dfrac1{2a}\log\left|\dfrac{a+x}{a-x}\right|+c$; the $x^2-a^2$ case runs identically, with the sign of one partial fraction flipped.

The third and fourth are proved instead by a trigonometric substitution. For $\int\frac{dx}{a^2+x^2}$, put $x=a\tan\theta$, so $dx=a\sec^2\theta\,d\theta$ and $a^2+x^2=a^2\sec^2\theta$; the integral collapses to $\frac1a\int d\theta=\frac1a\theta+c=\frac1a\tan^{-1}\!\left(\frac xa\right)+c$. For $\int\frac{dx}{\sqrt{a^2-x^2}}$, put $x=a\sin\theta$, so $dx=a\cos\theta\,d\theta$ and $a^2-x^2=a^2\cos^2\theta$; the $a\cos\theta$ in the numerator cancels the square root exactly, leaving $\int d\theta=\theta+c=\sin^{-1}\!\left(\frac xa\right)+c$. The remaining two (via $x=a\sec\theta$) work the same way, both ending in $\int\sec\theta\,d\theta=\log|\sec\theta+\tan\theta|+c$, which is converted back to $x$ using a right triangle with hypotenuse $x$ and one leg $a$; absorbing the leftover $-\log a$ into the constant of integration gives the tidy form $\log\left|x+\sqrt{x^2\mp a^2}\right|+c$.

A companion memory aid runs through the whole chapter: whenever the surd is $a^2-x^2$, substitute $x=a\sin\theta$; for $a^2+x^2$, substitute $x=a\tan\theta$; for $x^2-a^2$, substitute $x=a\sec\theta$.

Type II. Integrals $\int\frac{dx}{ax^2+bx+c}$ and $\int\frac{dx}{\sqrt{ax^2+bx+c}}$ are handled by first forcing the coefficient of $x^2$ to $1$ and then completing the square:

Whatever constant is left over after completing the square plays the role of "$a^2$" (or "$-a^2$") in a Type I form, in the shifted variable $x+\frac b{2a}$ -- so every Type II integral reduces directly to one of the six Type I results. The only genuinely new skill is the algebra of completing the square (and, for the square-root case with a leading coefficient $\ne1$, factoring that coefficient out from under the radical first).

Type III. Integrals with a linear numerator, $\int\frac{px+q}{ax^2+bx+c}\,dx$ and $\int\frac{px+q}{\sqrt{ax^2+bx+c}}\,dx$, are solved by writing the numerator as a multiple of the denominator's derivative, plus a leftover constant:

and finding $A$ and $B$ by comparing coefficients of $x$ and of the constant term on both sides. Substituting back splits the integral into two pieces:

The first piece is an $\int\frac{f'(x)}{f(x)}\,dx$ form and integrates directly to $A\log|ax^2+bx+c|$; the second piece is exactly a Type II integral, evaluated by completing the square. The square-root version runs in parallel: its first piece is an $\int f'(x)[f(x)]^{n}\,dx$ form with $n=-\tfrac12$, giving $2A\sqrt{ax^2+bx+c}$, while its second piece is a Type II square-root integral.

Remark (substitution table). As an alternative route to the same integrals, once the quadratic surd is written as $a^2-x^2$, $a^2+x^2$, or $x^2-a^2$ (after completing the square), it can also be attacked directly with the matching trigonometric substitution: $x=a\sin\theta$ for $a^2-x^2$, $x=a\tan\theta$ for $a^2+x^2$, and $x=a\sec\theta$ for $x^2-a^2$ -- the very substitutions used to prove the Type I square-root results.

Type IV. Integrals of the surd itself (not one over the surd), $\int\sqrt{a^2-x^2}\,dx$, $\int\sqrt{x^2-a^2}\,dx$, and $\int\sqrt{x^2+a^2}\,dx$, form Result 11.3:

$\int\sqrt{x^2-a^2}\,dx=\frac x2\sqrt{x^2-a^2}-\frac{a^2}2\log\left|x+\sqrt{x^2-a^2}\right|+c,$ …