8.2. Integrals using MF26 formulas

Illustrative examples:

$\displaystyle \int \frac{1}{x^2+4} \; \mathrm{d}x = \frac{1}{2} \tan^{-1} \left ( \frac{x}{2} \right) + C$

$\displaystyle \int \frac{1}{\sqrt{5-x^2}} \; \mathrm{d}x = \sin^{-1} \left ( \frac{x}{\sqrt{5}} \right ) + C$

$\displaystyle \int \frac{1}{9-4x^2} ; \mathrm{d}x = \frac{1}{12} \ln \left | \frac{3+2x}{3-2x} \right | + C$

$\displaystyle \int \frac{1}{x^2+2x-4} \; \mathrm{d}x = \frac{1}{2\sqrt{5}} \ln \left | \frac{x+1-\sqrt{5}}{x+1+\sqrt{5}} \right | + C$

The formulas

The following formulas will be provided in List MF26 for our examinations. $\displaystyle \int \frac{1}{x^2+a^2} \; \mathrm{d}x = \frac{1}{a} \tan^{-1} \left ( \frac{x}{a} \right ) \\ \displaystyle \int \frac{1}{\sqrt{a^2 -x^2}} \; \mathrm{d}x = \sin^{-1} \left ( \frac{x}{a} \right ) \qquad \mathrm{for} \; |x| < a \\ \displaystyle \int \frac{1}{x^2-a^2} \; \mathrm{d}x = \frac{1}{2a} \ln \left ( \frac{x-a}{x+1} \right ) \qquad \mathrm{for} \; x > a \\ \displaystyle \int \frac{1}{a^2-x^2} \; \mathrm{d}x = \frac{1}{2a} \ln \left ( \frac{a+x}{a-x} \right ) \qquad \mathrm{for} \; |x| < a$

For the last two formulas, note that $\displaystyle \int \frac{1}{x^2 - a^2} \; \mathrm{d}x = \displaystyle - \int \frac{1}{a^2 - x^2} \; \mathrm{d}x$ so it may seem a bit redundant to give 2 separate formulas for the same thing. The difference, however, lie in the domain. $\displaystyle \int \frac{1}{x^2-a^2}$ is valid for $x>a$ while the other integral is for $ |x| < a$.

We sometimes do not pay too much attention to the domain when integrating. This problem can be salvaged by introducing a modulus: both $\displaystyle \int \frac{1}{x^2 - a^2} \; \mathrm{d}x = \frac{1}{2a} \ln \left | \frac{x-a}{x+a} \right | +C$ and $\displaystyle \int \frac{1}{a^2 - x^2} \mathrm{d}x = \frac{1}{2a} \ln \left | \frac{a+x}{a-x} \right | + C$ work for all connected intervals within $\{x \in \mathbb{R}: x \neq a\}$.

Also note that the first 2 example questions can be solved by a direct application of the first 2 formulas.

Beware of coefficients

A common mistake is to get an answer of $\displaystyle \frac{1}{6} \ln \left | \frac{3+2x}{3-2x} \right |$ for the third example. Such an answer may seem intuitive: we let $a$ be $3$, and replace $x$ with $2x$ in the formula for $\displaystyle \int \frac{1}{a^2-x^2} \; \mathrm{d}x$. However, the problem, as seen in the discussion in example 8.0 , lies in the coefficient of $x$.

Method 1: Divide by the coefficient

To get to the correct answer, one method is to apply the reverse of chain rule and divide by the coefficient of $x$ in the last step during integration. Such an approach gives us $\displaystyle \int \frac{1}{9-4x^2} \; \mathrm{d}x = \int \frac{1}{3^2-(2x)^2} \; \mathrm{d}x = \frac{1}{2} \cdot \frac{1}{2(3)} \left ( \ln \left | \frac{3+2x}{3-2x} \right | \right ) + C.$ This leaves us with the answer $\displaystyle \frac{1}{12} \ln \left | \frac{3+2x}{3-2x} \right | + C$

Method 2: Force the coefficient of $x$ to be 1 by factorizing

If we are prone to forget to divide by the coefficient as in method 1, we can consider a second method: factorizing any coefficient we see before we integrate. This gives us $\displaystyle \int \frac{1}{9-4x^2} \; \mathrm{d}x = \int \frac{1}{4 \left ( \frac{9}{4} - x^2 \right )} \; \mathrm{d}x = \frac{1}{4} \cdot \frac{1}{2 \left ( \frac{3}{2} \right ) } \ln \left | \frac{\frac{3}{2}+x}{\frac{3}{2}-x} \right | + C $. Simplification gives us the same answer as in method 1.

Completing the square

For the last example, there is an extra "$2x$" term in our denominator such that we cannot apply the given formulas directly; the formulas involve perfect squares. Thankfully, we have a method to get perfect squares from any quadratic: completing the square.

$ \displaystyle \int \frac{1}{x^2 + 2x - 4} \; \mathrm{d}x = \int \frac{1}{(x+1)^2 - 1^2 - 4} \; \mathrm{d}x = \int \frac{1}{(x+1)^2 - 5} \; \mathrm{d}x$. Application of our formula gives us the answer $\displaystyle \frac{1}{2\sqrt{5}} \ln \left | \frac{x+1-\sqrt{5}}{x+1+\sqrt{5}} \right | + C$