Integration Reduced to Special Forms (Part 1)

This lesson handles two special types of integral: $\displaystyle\int \frac{dx}{ax^2 + bx + c}$ and $\displaystyle\int \frac{dx}{\sqrt{ax^2 + bx + c}}$. In both we complete the square on the quadratic to reach a standard form, then integrate.

The method: completing the square

Given $ax^2 + bx + c$:

1. If the coefficient of $x^2$ is not $1$, take it outside so the bracket starts with $x^2$.
2. Add and subtract the square of half the coefficient of $x$, that is $\left(\tfrac{b}{2}\right)^2$.
3. Group the first three terms as a perfect square $(x \pm k)^2$, leaving a constant behind.

This turns the quadratic into one of the standard forms below, which we then integrate directly.

Example 1: $\displaystyle\int \frac{dx}{x^2 + 6x + 13}$

The coefficient of $x^2$ is already $1$. Half of $6$ is $3$, and $3^2 = 9$, so add and subtract $9$:

$x^2 + 6x + 13 = (x^2 + 6x + 9) - 9 + 13 = (x + 3)^2 + 4 = (x + 3)^2 + 2^2.$

This is the standard form $\displaystyle\int \frac{dx}{x^2 + a^2} = \frac{1}{a}\tan^{-1}\frac{x}{a}$ with $a = 2$ and the variable $x + 3$:

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

Example 2: $\displaystyle\int \frac{dx}{9x^2 + 6x + 5}$

The coefficient of $x^2$ is $9$, so take $9$ outside:

$\int \frac{dx}{9x^2 + 6x + 5} = \frac{1}{9}\int \frac{dx}{x^2 + \tfrac{2}{3}x + \tfrac{5}{9}}.$

Half of $\tfrac{2}{3}$ is $\tfrac{1}{3}$, and $\left(\tfrac{1}{3}\right)^2 = \tfrac{1}{9}$. Add and subtract it:

$x^2 + \tfrac{2}{3}x + \tfrac{5}{9} = \left(x + \tfrac{1}{3}\right)^2 - \tfrac{1}{9} + \tfrac{5}{9} = \left(x + \tfrac{1}{3}\right)^2 + \tfrac{4}{9} = \left(x + \tfrac{1}{3}\right)^2 + \left(\tfrac{2}{3}\right)^2.$

Applying $\displaystyle\int \frac{dx}{x^2 + a^2} = \frac{1}{a}\tan^{-1}\frac{x}{a}$ with $a = \tfrac{2}{3}$:

$\frac{1}{9} \cdot \frac{1}{\tfrac{2}{3}} \tan^{-1}\frac{x + \tfrac{1}{3}}{\tfrac{2}{3}} = \frac{1}{9} \cdot \frac{3}{2}\tan^{-1}\frac{3x + 1}{2} = \frac{1}{6}\tan^{-1}\frac{3x + 1}{2} + C.$

Example 3: $\displaystyle\int \frac{dx}{\sqrt{7 - 6x - x^2}}$

Here the coefficient of $x^2$ inside the root is negative, so take $-1$ outside:

$7 - 6x - x^2 = -\left(x^2 + 6x - 7\right).$

Complete the square on $x^2 + 6x - 7$ using $3^2 = 9$:

$x^2 + 6x - 7 = (x + 3)^2 - 9 - 7 = (x + 3)^2 - 16.$

So

$7 - 6x - x^2 = -\left[(x + 3)^2 - 16\right] = 16 - (x + 3)^2 = 4^2 - (x + 3)^2.$

This is the standard form $\displaystyle\int \frac{dx}{\sqrt{a^2 - x^2}} = \sin^{-1}\frac{x}{a}$ with $a = 4$:

$\int \frac{dx}{\sqrt{4^2 - (x + 3)^2}} = \sin^{-1}\frac{x + 3}{4} + C.$

Example 4: $\displaystyle\int \frac{dx}{\sqrt{8 + 3x - x^2}}$

Again take $-1$ outside:

$8 + 3x - x^2 = -\left(x^2 - 3x - 8\right).$

Half of $3$ is $\tfrac{3}{2}$, and $\left(\tfrac{3}{2}\right)^2 = \tfrac{9}{4}$:

$x^2 - 3x - 8 = \left(x - \tfrac{3}{2}\right)^2 - \tfrac{9}{4} - 8 = \left(x - \tfrac{3}{2}\right)^2 - \tfrac{41}{4},$

since $-\tfrac{9}{4} - 8 = \dfrac{-9 - 32}{4} = -\tfrac{41}{4}$. Therefore

$8 + 3x - x^2 = \tfrac{41}{4} - \left(x - \tfrac{3}{2}\right)^2 = \left(\tfrac{\sqrt{41}}{2}\right)^2 - \left(x - \tfrac{3}{2}\right)^2.$

Using $\displaystyle\int \frac{dx}{\sqrt{a^2 - x^2}} = \sin^{-1}\frac{x}{a}$ with $a = \tfrac{\sqrt{41}}{2}$:

$\sin^{-1}\frac{x - \tfrac{3}{2}}{\tfrac{\sqrt{41}}{2}} = \sin^{-1}\frac{2x - 3}{\sqrt{41}} + C.$

Example 5: $\displaystyle\int \frac{dx}{\sqrt{5x^2 - 2x}}$

The coefficient of $x^2$ is $5$, so take it outside the root as $\sqrt{5}$:

$\int \frac{dx}{\sqrt{5x^2 - 2x}} = \frac{1}{\sqrt{5}}\int \frac{dx}{\sqrt{x^2 - \tfrac{2}{5}x}}.$

Half of $\tfrac{2}{5}$ is $\tfrac{1}{5}$, and $\left(\tfrac{1}{5}\right)^2 = \tfrac{1}{25}$:

$x^2 - \tfrac{2}{5}x = \left(x - \tfrac{1}{5}\right)^2 - \tfrac{1}{25} = \left(x - \tfrac{1}{5}\right)^2 - \left(\tfrac{1}{5}\right)^2.$

This is the standard form $\displaystyle\int \frac{dx}{\sqrt{x^2 - a^2}} = \log\left|x + \sqrt{x^2 - a^2}\right|$ with variable $x - \tfrac{1}{5}$ and $a = \tfrac{1}{5}$:

$\frac{1}{\sqrt{5}}\log\left|\left(x - \tfrac{1}{5}\right) + \sqrt{x^2 - \tfrac{2}{5}x}\right| + C.$

Note that the square root term keeps the reduced expression $\sqrt{x^2 - \tfrac{2}{5}x}$, because the factor $\sqrt{5}$ was already taken outside.

Key takeaways

• Complete the square first: $ax^2 + bx + c$ becomes $(x \pm k)^2 \pm m$, and a negative $x^2$ term becomes $a^2 - (x \pm k)^2$ after taking $-1$ outside.
• A quadratic denominator gives an inverse tangent; a square root of the form $a^2 - x^2$ gives an inverse sine; a square root of the form $x^2 - a^2$ gives a logarithm.
• When the coefficient of $x^2$ is not $1$, take it outside first, and remember it stays as $\sqrt{\cdot}$ outside the root in the square-root cases.