5.18 More integrals

1. Integration by Substitution

When we need to integrate a composite function where the derivative of the "inner" function is also present in the integrand, we can reverse the chain rule using integration by substitution.

$\displaystyle\int f(g(x))g'(x) dx = \int f(u) du$

By letting $u = g(x)$, we find the differential $du = g'(x)dx$, which seamlessly translates the integral entirely into terms of $u$.

Example 1

Evaluate the indefinite integral $\int 2x \sqrt{x^2 + 1} dx$.

Let $u = x^2 + 1$.
Then differentiating with respect to $x$ gives $\dfrac{du}{dx} = 2x \implies du = 2x dx$. We substitute $u$ and $du$ into the original integral $$ \int \sqrt{x^2 + 1} (2x dx) = \int \sqrt{u} du = \int u^{1/2} du = \dfrac{u^{3/2}}{3/2} + c = \dfrac{2}{3} u^{3/2} + c $$ Finally, we replace $u$ with the original expression $x^2 + 1$,
$$ \int 2x \sqrt{x^2 + 1} dx = \dfrac{2}{3} (x^2 + 1)^{3/2} + c $$

2. Integrals Yielding Inverse Yrigonometric Functions HL ONLY

The IB formula booklet provides standard integrals that result in inverse trigonometric functions. It is worthwhile to memorize their structure so you can efficiently spot them.

  • $\displaystyle\int \dfrac{1}{\sqrt{a^2 - x^2}} dx = \arcsin\left(\dfrac{x}{a}\right) + c \quad (|x| < a)$
  • $\displaystyle\int \dfrac{1}{a^2 + x^2} dx = \dfrac{1}{a} \arctan\left(\dfrac{x}{a}\right) + c$

Example 2

Evaluate the integral $\displaystyle\int \dfrac{1}{x^2 + 9} dx$.

Notice that the denominator takes the form $x^2 + a^2$. Here, $a^2 = 9 \implies a = 3$. Applying the standard formula for $\arctan$ directly yields
$$ \int \dfrac{1}{x^2 + 3^2} dx = \mathbf{\dfrac{1}{3} \arctan\left(\dfrac{x}{3}\right) + c} $$

3. Integration Using Partial Fractions HL ONLY

When integrating a rational function where the denominator can be factored into distinct linear polynomials, we decompose the fraction into simpler partial fractions. This strictly converts a difficult fraction into basic logarithmic integrals.

Example 3

Evaluate the integral $\displaystyle\int \dfrac{2}{x^2 - 1} dx$.

First, factor the denominator: $x^2 - 1 = (x - 1)(x + 1)$.
Next, set up the partial fraction decomposition $$ \dfrac{2}{(x - 1)(x + 1)} = \dfrac{A}{x - 1} + \dfrac{B}{x + 1} $$ Multiply through by the common denominator to equate the numerators $$ 2 = A(x + 1) + B(x - 1) $$ Let $x = 1 \implies 2 = 2A \implies A = 1$
Let $x = -1 \implies 2 = -2B \implies B = -1$
Now, substitute these constants back into the integral $$ \begin{aligned} \int \dfrac{2}{x^2 - 1} dx &= \int \left( \dfrac{1}{x - 1} - \dfrac{1}{x + 1} \right) dx \\ &= \int \dfrac{1}{x - 1} dx - \int \dfrac{1}{x + 1} dx \\ &= \ln|x - 1| - \ln|x + 1| + c \\ &= \mathbf{\ln\left|\dfrac{x - 1}{x + 1}\right| + c} \end{aligned} $$