Indefinite integrals

This page is about evaluating indefinite integrals and using them to solve simple differential equations.

Antidifferentiation
Common antiderivatives
Properties
Linear over linear integration
Simple differential equations
Notes on absolute values

Antidifferentiation

A property of indefinite integrals is that they are antiderivatives, which “undo” differentiation.

Eg. $\displaystyle \int\frac{1}{x} \d x = \ln\lvert x\rvert + C$ , where $C$ is any constant of integration. Here $\displaystyle \frac{1}{x}$ is the integrand and $\ln\lvert x\rvert + C$ is the antiderivative or indefinite integral.

Usually, the $+C$ is worth one mark!

The $+C$ indicates that antiderivatives are not unique, as many answers are returned off by a constant, any constant.

There is also the notation

\int \frac{\d x}{f(x)} = \int \frac{1}{f(x)}\d x

Common antiderivatives

Reading the tables in Common derivatives and More derivatives (HL) in reverse yields common antiderivatives.

Here are common antiderivatives you should know

function	antiderivatives
$k$ , not a function of $x$	$kx + C$
$x^n, n \neq -1$	$\displaystyle \frac{1}{n+1}x^{n+1} + C$
$x^{-1}$	$\ln\lvert x\rvert + C$
$\e^x$	$\e^x + C$
$\sin x$	$-\cos x + C$
$\cos x$	$\sin x + C$

The absolute value in $\ln\lvert x\rvert + C$ aims to give the antiderivatives the same domain as the integrand.

Also remember

\cos2x = 2\cos^2x-1 = 1-2\sin^2x

or that

\cos^2x = \frac{1 + \cos2x}{2}

\sin^2x = \frac{1 - \cos2x}{2}

for integrating $\cos^2 x$ and $\sin^2 x$ .

At HL, the formula booklet also lists the following integrals

function	antiderivatives
$a^x$	$\displaystyle \frac1{\ln a} a^x + C$
$\displaystyle \frac{1}{a^2 + x^2}$	$\displaystyle \frac1a \arctan\left(\frac xa\right) + C$
$\displaystyle \frac{1}{\sqrt{a^2 - x^2}}$	$\displaystyle \arcsin \left(\frac xa\right) + C,\, \lvert x\rvert < \lvert a\rvert$

Properties

Unlike differentiation, there lack formulas for antiderivatives of composite functions, or product or quotient of functions, because there lack differentiation formulas resulting in such functions.

Instead, analogous properties involving constants are presented below. $F(x)$ is an antiderivative of $f(x)$ .

\int f(x) + g(x)\d x = \int f(x) \d x + \int g(x) \d x

\int k f(x)\d x = kF(x) + C

\int f(ax + b)\d x = \frac 1a \int f(x)\d x = \frac 1aF(ax + b) + C

The last one is a special case of integration by substitution.

Continuous functions are always integrable (meaning antiderivatives exist), even if we often cannot easily evaluate them.

Example: Simplify

\int \left(\sqrt[3]{4x+1} + \frac{1}{\sqrt[3]{4x+1}}\right)^2 \d x

Rewrite radicals (surds) as fractional exponents. Then expand using ${(a+b)^2 = a^2 = 2ab + b^2}$

\begin{align*} \text{integrand} &= \left((4x+1)^{\frac13} + (4x+1)^{-\frac13}\right)^2 \\ &= (4x+1)^{\frac23} + 2 + (4x+1)^{-\frac23} \\ \end{align*}

Now, use ${\int x^n \d x = \frac{1}{n+1}x^{n+1} + C}$ and ${\int f(ax + b) = \frac1a F(ax + b) + C}$ to integrate term by term.

\begin{align*} \int (4x+1)^{\frac23} \d x &= \frac1{4\left(\frac23 + 1\right)} (4x + 1)^{\frac23 + 1} + C_1 \\ &= \frac{3}{20}(4x + 1)^{\frac53} + C_1 \end{align*}

\begin{align*} \int 2 \d x &= 2x + C_2 \\ \end{align*}

\begin{align*} \int (4x+1)^{-\frac23} \d x &= \frac1{4\left(-\frac23 + 1\right)} (4x + 1)^{-\frac23 + 1} + C_3 \\ &= \frac{3}{4}(4x + 1)^{\frac13} + C_3 \end{align*}

Putting everything together

\begin{align*} I &= \int \left(\sqrt[3]{4x+1} + \frac{1}{\sqrt[3]{4x+1}}\right)^2 \d x \\ &= \frac{3}{20}(4x + 1)^{\frac53} + 2x + \frac{3}{4}(4x + 1)^{\frac13} + C \qed \end{align*}

The constants $C_1$ , $C_2$ and $C_3$ were collected into $C$ .

Note that we used the real root, as opposed to the complex principal root [HL] when interpreting $\sqrt[3]{4x+1}$ . This is typically the case as in this course we only work with real functions.

Linear over linear integration

There is a small chance that all candidates may be asked to integrate a linear over linear rational function.

The general strategy is to convert it to a $\text{constant } + \text{rational}$ form.

Example: Simplify

\int \frac{4x + 1}{2x + 3} \d x

Because $2(2x+3) = 4x + 6$ and $4x + 6 - 5 = 4x + 1$ , then

\begin{align*} \frac{4x + 1}{2x + 3} &= \frac{2(2x+3) - 5}{2x+3} \\ &= 2 - \frac{5}{2x + 3} \end{align*}

\begin{align*} \int 2 - \frac{5}{2x + 3} \d x &= 2x - \frac{5}{2} \ln\lvert2x+3\rvert + C\qed \end{align*}

where we invoked the property of dividing by $\frac{1}{a}$ when integrating $f(ax + b)$ .

For integrating linear over quadratic functions, HL candidates should refer to rational functions integration.

Simple differential equations

With antiderivatives, you can solve differential equations of the forms

\frac{\d f}{\d x} = g(x)

\frac{\d ^2f}{\d x^2} = g(x)

As well as at HL,

\frac{\d ^n}{\d x^n} f(x) = g(x)

This means, given the $1$ st, $2$ nd, or $n$ th derivative as $g(x)$ , find the original function $f(x)$ .

Note that this technique cannot be used if there are multiple derivative functions, ie both $\frac{\d f}{\d x}$ and $\frac{\d ^2f}{\d x^2}$ in the same equation. The technique also fails when there are variables on the right side other than the variable differentiating with respect to.

Examples of differential equations that this could solve

\frac{\d y}{\d x} = 2x

\frac{\d ^2x}{\d t^2} = \e^{-3t}

Examples of differential equations that could not be solved this way

\frac{\d m}{\d x} = m

\frac{\d ^2x}{\d t^2} + 2\frac{\d y}{\d x} = t \e^{-3x}

The solution to a differential equation is a function. In addition, $n$ boundary conditions, one for each level of differentiation, should be given to pin down the answer to a specific function.

Example: The position of a particle in free fall satisfies the equation

\frac{\d ^2y}{\d t^2} = -9.8

where $y$ is the height above Earth’s surface.

Given $\displaystyle y^\prime(0) = u$ and $y(0) = y_0$ . Find $y(t)$ while the particle experiences gravity.

Integrating both sides once gives

y^\prime(t) = -9.8t + C

using $\displaystyle y^\prime(0) = u$ results in

\begin{align*}y^\prime(0) &= -9.8(0) + C = u \\ C &= u \\ y^\prime(t)&= -9.8t + u \end{align*}

Integrate again gives

y(t) = -4.9t^2 + ut + D

Using a different constant of integration as it’s a different integral in the same problem. Using $y(0) = y_0$ we get

\begin{align*}y(0) &= -4.9(0)^2 + u(0) + D = y_0 \\ D &= y_0 \\ y(t) &= -4.9t^2 + ut + y_0 \qed \end{align*}

Notes on absolute values

If the integrand contains absolute value $\lvert f(x)\rvert$ , then you must find a way to re-express the integrand without absolute values, probably after some simplification first. You may need to find the zeros, and break into domains of positive or negative $f(x)$ , or use some property of the specific functions used. Do not try to invent absolute value function integration rules.

For integrating factor (HL), where applicable, use

\e^{\int \frac1x \d x} = \e^{\ln x} = x