Integration Reduced to Special Forms (Part 2)

This lesson shows how to integrate a linear expression divided by a quadratic, and a linear expression divided by the square root of a quadratic. The key idea is to rewrite the numerator $Px + Q$ as a constant multiple of the derivative of the quadratic, plus a second constant.

The general method

For an integral of the form

$\int \frac{Px + Q}{ax^2 + bx + c}\,dx \qquad\text{or}\qquad \int \frac{Px + Q}{\sqrt{ax^2 + bx + c}}\,dx$

write the numerator as

$Px + Q = A\cdot\frac{d}{dx}\bigl(ax^2 + bx + c\bigr) + B = A\,(2ax + b) + B.$

Here $A$ and $B$ are real constants. Compare the coefficients of $x$ and the constant terms on both sides to find $A$ and $B$, substitute them back, and integrate the two resulting pieces.

Example 1: a linear term over a quadratic

$I = \int \frac{5x + 2}{3x^2 + 2x + 1}\,dx.$

Write the numerator using the derivative of the denominator, $\dfrac{d}{dx}(3x^2 + 2x + 1) = 6x + 2$:

$5x + 2 = A\,(6x + 2) + B.$

Comparing coefficients of $x$: $\;6A = 5$, so $A = \tfrac{5}{6}$.

Comparing constants: $\;2A + B = 2$, so $\tfrac{5}{3} + B = 2$ and $B = \tfrac{1}{3}$.

The integral splits into two parts:

$I = \tfrac{5}{6}\int \frac{6x + 2}{3x^2 + 2x + 1}\,dx + \tfrac{1}{3}\int \frac{1}{3x^2 + 2x + 1}\,dx = I_1 + I_2.$

First part

The numerator $6x + 2$ is exactly the derivative of the denominator, so this is of the form $\int \frac{f'(x)}{f(x)}\,dx = \log|f(x)|$:

$I_1 = \tfrac{5}{6}\,\log\bigl|3x^2 + 2x + 1\bigr|.$

Second part

Take out the leading coefficient and complete the square:

$I_2 = \tfrac{1}{3}\int \frac{dx}{3\left(x^2 + \tfrac{2}{3}x + \tfrac{1}{3}\right)} = \tfrac{1}{9}\int \frac{dx}{\left(x + \tfrac{1}{3}\right)^2 + \tfrac{2}{9}}.$

With $\left(\tfrac{\sqrt{2}}{3}\right)^2 = \tfrac{2}{9}$, use the standard result $\int \frac{dx}{x^2 + a^2} = \tfrac{1}{a}\tan^{-1}\tfrac{x}{a}$:

$I_2 = \tfrac{1}{9}\cdot \frac{3}{\sqrt{2}}\,\tan^{-1}\!\left(\frac{x + \tfrac{1}{3}}{\tfrac{\sqrt{2}}{3}}\right) = \frac{1}{3\sqrt{2}}\,\tan^{-1}\!\left(\frac{3x + 1}{\sqrt{2}}\right).$

Combining,

$I = \tfrac{5}{6}\,\log\bigl|3x^2 + 2x + 1\bigr| + \frac{1}{3\sqrt{2}}\,\tan^{-1}\!\left(\frac{3x + 1}{\sqrt{2}}\right) + C.$

Example 2: a linear term over a square root

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

Using $\dfrac{d}{dx}(4x - x^2) = 4 - 2x$, write

$x + 2 = A\,(4 - 2x) + B.$

Comparing coefficients of $x$: $\;-2A = 1$, so $A = -\tfrac{1}{2}$.

Comparing constants: $\;4A + B = 2$, so $-2 + B = 2$ and $B = 4$.

The integral splits as

$I = -\tfrac{1}{2}\int \frac{4 - 2x}{\sqrt{4x - x^2}}\,dx + 4\int \frac{dx}{\sqrt{4x - x^2}} = I_1 + I_2.$

First part

Put $t = 4x - x^2$, so $dt = (4 - 2x)\,dx$:

$I_1 = -\tfrac{1}{2}\int \frac{dt}{\sqrt{t}} = -\tfrac{1}{2}\cdot 2\sqrt{t} = -\sqrt{4x - x^2}.$

Second part

Complete the square under the root, $4x - x^2 = 4 - (x - 2)^2$, and use $\int \frac{dx}{\sqrt{a^2 - x^2}} = \sin^{-1}\tfrac{x}{a}$:

$I_2 = 4\int \frac{dx}{\sqrt{2^2 - (x - 2)^2}} = 4\,\sin^{-1}\!\left(\frac{x - 2}{2}\right).$

Combining,

$I = -\sqrt{4x - x^2} + 4\,\sin^{-1}\!\left(\frac{x - 2}{2}\right) + C.$

Key takeaways

• Write the linear numerator as $A$ times the derivative of the quadratic plus a constant $B$, then find $A$ and $B$ by comparing coefficients.
• The derivative part integrates to a logarithm (for a quadratic denominator) or a square root (for a square-root denominator).
• The leftover constant part is finished by completing the square, giving an inverse tangent or an inverse sine.