Integration: Special Forms (Part 1)

This lesson works through integrals of special forms, applying the standard formulas from Exercise 7.4. Each integral is reshaped to match a known pattern, sometimes after a substitution, and then read off using the corresponding result.

One over a difference of squares

Given $\displaystyle\int \frac{dx}{x^2 - 16}$, write $16 = 4^2$ so it matches

$\int \frac{dx}{x^2 - a^2} = \frac{1}{2a}\log\left|\frac{x-a}{x+a}\right| + C.$

With $a = 4$:

$\int \frac{dx}{x^2 - 16} = \frac{1}{8}\log\left|\frac{x-4}{x+4}\right| + C.$

A substitution giving an inverse tangent

For $\displaystyle\int \frac{3x^2}{x^6 + 1}\,dx$, notice $x^6 = (x^3)^2$. Put $t = x^3$, so $dt = 3x^2\,dx$:

$\int \frac{dt}{t^2 + 1} = \tan^{-1} t + C = \tan^{-1}(x^3) + C.$

Square-root forms giving a logarithm

Using $\displaystyle\int \frac{dx}{\sqrt{a^2 + x^2}} = \log\left|x + \sqrt{x^2 + a^2}\right| + C$.

One over root of one plus four x squared

Write $4x^2 = (2x)^2$. With the inner variable $2x$, divide by its derivative $2$:

$\int \frac{dx}{\sqrt{1 + 4x^2}} = \frac{1}{2}\log\left|2x + \sqrt{1 + 4x^2}\right| + C.$

A shifted square root

For $\displaystyle\int \frac{dx}{\sqrt{(2-x)^2 + 1}}$, the inner variable is $2-x$ with derivative $-1$:

$\int \frac{dx}{\sqrt{(2-x)^2 + 1}} = -\log\left|(2-x) + \sqrt{(2-x)^2 + 1}\right| + C.$

Reducing to an inverse sine

Using $\displaystyle\int \frac{dx}{\sqrt{a^2 - x^2}} = \sin^{-1}\frac{x}{a} + C$.

For $\displaystyle\int \frac{dx}{\sqrt{9 - 25x^2}}$, take $25$ outside the root, leaving $5\sqrt{\tfrac{9}{25} - x^2}$, so $a = \tfrac{3}{5}$:

$\int \frac{dx}{\sqrt{9 - 25x^2}} = \frac{1}{5}\sin^{-1}\frac{x}{3/5} + C = \frac{1}{5}\sin^{-1}\frac{5x}{3} + C.$

Substitution before an inverse tangent

For $\displaystyle\int \frac{3x}{1 + 2x^4}\,dx$, put $t = x^2$, so $x\,dx = \tfrac{1}{2}\,dt$:

$\frac{3}{2}\int \frac{dt}{1 + 2t^2}.$

Matching $\int \frac{dt}{a^2 + t^2} = \frac{1}{a}\tan^{-1}\frac{t}{a}$ with $a = \tfrac{1}{\sqrt 2}$:

$\int \frac{3x}{1 + 2x^4}\,dx = \frac{3\sqrt 2}{4}\tan^{-1}\!\left(\sqrt 2\,x^2\right) + C.$

A logarithm of a quotient

For $\displaystyle\int \frac{x^2}{1 - x^6}\,dx$, put $t = x^3$, so $x^2\,dx = \tfrac{1}{3}\,dt$:

$\frac{1}{3}\int \frac{dt}{1 - t^2} = \frac{1}{3}\cdot\frac{1}{2}\log\left|\frac{1+t}{1-t}\right| = \frac{1}{6}\log\left|\frac{1+x^3}{1-x^3}\right| + C.$

Splitting a numerator

For $\displaystyle\int \frac{x-1}{\sqrt{x^2 - 1}}\,dx$, split into two integrals $I = I_1 - I_2$.

$I_1$. For $\displaystyle\int \frac{x}{\sqrt{x^2 - 1}}\,dx$, put $t = x^2 - 1$, so $x\,dx = \tfrac{1}{2}\,dt$:

$\frac{1}{2}\int \frac{dt}{\sqrt t} = \sqrt t = \sqrt{x^2 - 1}.$

$I_2$. Using $\displaystyle\int \frac{dx}{\sqrt{x^2 - a^2}} = \log\left|x + \sqrt{x^2 - a^2}\right|$ with $a = 1$:

$I_2 = \log\left|x + \sqrt{x^2 - 1}\right|.$

Combining:

$\int \frac{x-1}{\sqrt{x^2 - 1}}\,dx = \sqrt{x^2 - 1} - \log\left|x + \sqrt{x^2 - 1}\right| + C.$

Another root substitution

For $\displaystyle\int \frac{x^2}{\sqrt{x^6 + a^6}}\,dx$, put $t = x^3$, so $x^2\,dx = \tfrac{1}{3}\,dt$, and $a^6 = (a^3)^2$:

$\frac{1}{3}\int \frac{dt}{\sqrt{t^2 + (a^3)^2}} = \frac{1}{3}\log\left|x^3 + \sqrt{x^6 + a^6}\right| + C.$

Square-root integrals with a half-product

Using $\displaystyle\int \sqrt{a^2 - x^2}\,dx = \frac{x}{2}\sqrt{a^2 - x^2} + \frac{a^2}{2}\sin^{-1}\frac{x}{a} + C$.

Root of four minus x squared

With $a = 2$:

$\int \sqrt{4 - x^2}\,dx = \frac{x}{2}\sqrt{4 - x^2} + 2\sin^{-1}\frac{x}{2} + C.$

Root of one plus x squared over nine

Write $\sqrt{1 + \tfrac{x^2}{9}} = \tfrac{1}{3}\sqrt{9 + x^2}$ and use $\displaystyle\int \sqrt{a^2 + x^2}\,dx = \frac{x}{2}\sqrt{a^2 + x^2} + \frac{a^2}{2}\log\left|x + \sqrt{a^2 + x^2}\right|$ with $a = 3$:

$\int \sqrt{1 + \tfrac{x^2}{9}}\,dx = \frac{x}{6}\sqrt{9 + x^2} + \frac{3}{2}\log\left|x + \sqrt{9 + x^2}\right| + C.$

Key takeaways

• Reshape each integral to match a standard special-form formula, taking constants outside the root or square so the pattern is clear.
• A substitution like $t = x^2$ or $t = x^3$ often turns a hard integral into a recognisable inverse-tangent, inverse-sine, or logarithm form.
• When the variable inside a formula carries a coefficient, divide by its derivative to keep the result correct.