How Six Standard Integrals Unlock Every Quadratic Denominator

Integrals like $\int \frac{dx}{x^2+4}$ or $\int \frac{dx}{\sqrt{9-x^2}}$ do not fit the power rule or a simple substitution, yet they appear constantly in areas, curve lengths and physics. A small set of standard results, together with completing the square, handles them all.

This lesson covers the basic standard forms, quadratic denominators and roots, linear numerators, and reducing an integral to a standard form.

**The standard results, each plus C, are $\int \frac{dx}{x^2+a^2}=\frac{1}{a}{\tan }^{-1}\frac{x}{a}$, $\int \frac{dx}{x^2-a^2}=\frac{1}{2a}\log \left|\frac{x-a}{x+a}\right|$, $\int \frac{dx}{a^2-x^2}=\frac{1}{2a}\log \left|\frac{a+x}{a-x}\right|$, $\int \frac{dx}{\sqrt{a^2-x^2}}={\sin }^{-1}\frac{x}{a}$, and $\int \frac{dx}{\sqrt{x^2\pm a^2}}=\log |x+\sqrt{x^2\pm a^2}|$.

Where they come from.** Partial fractions give the logarithmic forms; the substitutions $x=a\tan \theta$ and $x=a\sin \theta$ give the inverse tangent and inverse sine.

Worked examples:

- $\int \frac{dx}{x^2+9}=\frac{1}{3}{\tan }^{-1}\frac{x}{3}+C$
- $\int \frac{dx}{x^2-16}=\frac{1}{8}\log \left|\frac{x-4}{x+4}\right|+C$
- $\int \frac{dx}{\sqrt{25-x^2}}={\sin }^{-1}\frac{x}{5}+C$

Worked example 2 (scaling). $\int \frac{dx}{4x^2+1}=\frac{1}{4}\int \frac{dx}{x^2+\frac{1}{4}}=\frac{1}{4}\times 2{\tan }^{-1}(2x)+C=\frac{1}{2}{\tan }^{-1}(2x)+C$.

An everyday example. Working out how a streetlight's glow spreads along a road leads to $\int \frac{dx}{x^2+a^2}$, where a is the height of the lamp.

The substance. Match the sign pattern exactly — $x^2+a^2$ gives an inverse tangent, but $x^2-a^2$ gives a logarithm.

**Complete the square to write the quadratic as a constant times ${\left(x+\frac{b}{2a}\right)}^2$ plus or minus a square, then apply the matching standard form with $x+\frac{b}{2a}$ in place of x.

Worked example.** Find $\int \frac{dx}{x^2+6x+13}$.

$$x^2+6x+13=(x+3{)}^2+4\Rightarrow \int \frac{dx}{(x+3{)}^2+2^2}=\frac{1}{2}{\tan }^{-1}\frac{x+3}{2}+C$$

Worked example 2. Find $\int \frac{dx}{\sqrt{8+2x-x^2}}$.

$$8+2x-x^2=9-(x-1{)}^2\Rightarrow \int \frac{dx}{\sqrt{3^2-(x-1{)}^2}}={\sin }^{-1}\frac{x-1}{3}+C$$

Worked example 3. Find $\int \frac{dx}{2x^2+4x-6}$. Since $2x^2+4x-6=2[(x+1{)}^2-4]$, the integral is $\frac{1}{2}\times \frac{1}{4}\log \left|\frac{x-1}{x+3}\right|+C=\frac{1}{8}\log \left|\frac{x-1}{x+3}\right|+C$.

An everyday example. Rewriting a messy measurement like 2.75 cups as '3 cups minus a quarter cup' before measuring is the same tidy-up that completing the square gives a quadratic.

The substance. **Take out the coefficient of $x^2$ first** — forgetting the factor 2 in the last example changes the answer by that factor.

Write the numerator as A times the derivative of the quadratic plus a constant B, so the integral splits into a derivative-over-function part, giving a logarithm or a square root, and a constant part handled by the earlier standard forms.

Worked example. Find $\int \frac{3x+5}{x^2+2x+5}dx$.

- Write $3x+5=\frac{3}{2}(2x+2)+2$
- First part: $\frac{3}{2}\int \frac{2x+2}{x^2+2x+5}dx=\frac{3}{2}\log (x^2+2x+5)$
- Second part: $2\int \frac{dx}{(x+1{)}^2+4}={\tan }^{-1}\frac{x+1}{2}$

$$\int \frac{3x+5}{x^2+2x+5}dx=\frac{3}{2}\log (x^2+2x+5)+{\tan }^{-1}\frac{x+1}{2}+C$$

Worked example 2 (square root). $\int \frac{x+2}{\sqrt{x^2+4x+13}}dx=\frac{1}{2}\int \frac{2x+4}{\sqrt{x^2+4x+13}}dx=\sqrt{x^2+4x+13}+C$, because $\int \frac{f^{\prime }}{\sqrt{f}}dx=2\sqrt{f}$.

An everyday example. Splitting a phone bill into the part covered by a standard plan and a small extra charge mirrors splitting the numerator into a derivative part and a constant.

The substance. **Over a square root, the derivative part gives $2\sqrt{f}$, not a logarithm** — the logarithm appears only when the quadratic is not under a root.

**Reduce an integral with a substitution or rearrangement that exposes $x^2\pm a^2$, $a^2-x^2$ or a quadratic — for example $t=e^x$, $t=\sin x$ or $t=x^2$ — and then apply the matching standard form.

Worked example (exponential).** Find $\int \frac{e^x}{e^{2x}+1}dx$. With $t=e^x$ and $dt=e^{x}dx$:

$$\int \frac{dt}{t^2+1}={\tan }^{-1}t+C={\tan }^{-1}(e^x)+C$$

Worked example (trigonometric). Find $\int \frac{\cos x}{{\sin }^{2}x+4\sin x+5}dx$. With $t=\sin x$, the integral is $\int \frac{dt}{(t+2{)}^2+1}={\tan }^{-1}(\sin x+2)+C$.

Worked example (power). Find $\int \frac{x}{\sqrt{1-x^4}}dx$. With $t=x^2$ and $dt=2xdx$, it becomes $\frac{1}{2}\int \frac{dt}{\sqrt{1-t^2}}=\frac{1}{2}{\sin }^{-1}(x^2)+C$.

An everyday example. Recognising that an unfamiliar recipe is really a dish you already know saves starting from scratch — the substitution turns a new integral into a familiar form.

The substance. Look for a function paired with its derivative — $e^x$ beside $e^{2x}$, or $\cos x$ beside powers of $\sin x$, signals the substitution.