Standard Integral Forms with Algebraic Denominators

Two integrals recur throughout calculus and mechanics: $\displaystyle\int\frac{dx}{x^2+a^2}$

and $\displaystyle\int\frac{dx}{x^2-a^2}$ (the syllabus's $\int dx/(x^2\pm a^2)$). Both are

derived once here and then reused, via completing the square, for every quadratic denominator.

Derivation of $\int dx/(x^2+a^2)$ by trigonometric substitution. Let $x=a\tan\theta$, so

$dx=a\sec^2\theta\,d\theta$ and $x^2+a^2=a^2\tan^2\theta+a^2=a^2\sec^2\theta$ (using

$1+\tan^2\theta=\sec^2\theta$). Then

using $\theta=\tan^{-1}(x/a)$ from the substitution. This is verified directly by differentiating

the right side: $\dfrac{d}{dx}\!\left[\dfrac1a\tan^{-1}(x/a)\right] =\dfrac1a\cdot\dfrac{1/a}{1+x^2/a^2}=\dfrac{1}{x^2+a^2}$.

Derivation of $\int dx/(x^2-a^2)$ by partial fractions. Since $x^2-a^2=(x-a)(x+a)$,

$\dfrac{1}{x^2-a^2}=\dfrac{1}{2a}\!\left[\dfrac{1}{x-a}-\dfrac{1}{x+a}\right]$ (Section 3), so

Reducing a general quadratic denominator: complete the square. For

$\int dx/(ax^2+bx+c)$, first factor out $a$ and complete the square inside:

Writing $u=x+b/(2a)$ and $k^2=|4ac-b^2|/(4a^2)$, the bracket becomes $u^2+k^2$ (if

$4ac>b^2$) or $u^2-k^2$ (if $4ac<b^2$) -- exactly the two forms derived above, now in $u$, giving

a $\tan^{-1}$ or a $\ln$ answer respectively. Example 8 carries this out for

$\int dx/(x^2+4x+13)$, where $(x+2)^2+9$ gives the $\tan^{-1}$ case.

The linear-numerator form $\int(px+q)\,dx/(ax^2+bx+c)$. The denominator's derivative is …