Integration as the reverse of differentiation: antiderivatives and standard integrals

Unit 16 · Integration

Find an antiderivative by reversing differentiation, add the constant of integration, and use the standard integrals and the linearity rules, including $f(ax+b)$.

Reversing differentiation

Differentiation takes $x^3$ to $3x^2$. Integration asks the opposite question: given $f(x) = 3x^2$, which function has $f$ as its derivative? Any such function $F$, with $F'(x) = f(x)$, is called an antiderivative of $f$. Here $F(x) = x^3$ works, but so do $x^3 + 5$ and $x^3 - 7$, because a constant disappears when we differentiate. So one antiderivative always comes with a free constant, and we write $$\int f(x),dx = F(x) + c.$$ This is the indefinite integral of $f$. The symbol $\int$ is read "integral", $f(x)$ is the function being integrated, $dx$ shows the variable, and $c$ is the constant of integration. Never leave $c$ out of an indefinite integral.

You can always check an answer: differentiate it. If you get back the function you were integrating, it is correct.

Standard integrals

Reading each differentiation result backwards gives the standard integrals:

• $\displaystyle\int x^n,dx = \dfrac{x^{n+1}}{n+1} + c$ for $n \ne -1$ (raise the power by one and divide by the new power). This holds for negative and fractional $n$ too.
• $\displaystyle\int \dfrac{1}{x},dx = \ln|x| + c$ (the case $n = -1$). The modulus is needed because $\dfrac{d}{dx}\ln|x| = \dfrac{1}{x}$ for $x<0$ as well.
• $\displaystyle\int e^x,dx = e^x + c$, $\displaystyle\int a^x,dx = \dfrac{a^x}{\ln a} + c$.
• $\displaystyle\int \cos x,dx = \sin x + c$, $\displaystyle\int \sin x,dx = -\cos x + c$ (note the minus sign).
• $\displaystyle\int \sec^2 x,dx = \tan x + c$, $\displaystyle\int \operatorname{cosec}^2 x,dx = -\cot x + c$.
• $\displaystyle\int \sec x\tan x,dx = \sec x + c$, $\displaystyle\int \operatorname{cosec} x\cot x,dx = -\operatorname{cosec} x + c$.
• $\displaystyle\int \dfrac{dx}{1+x^2} = \tan^{-1}x + c$, $\displaystyle\int \dfrac{dx}{\sqrt{1-x^2}} = \sin^{-1}x + c$.

Rules: sums, constants and $f(ax+b)$

Because differentiation is linear, so is integration: $$\int [af(x) + bg(x)],dx = a\int f(x),dx + b\int g(x),dx.$$ So we integrate a sum term by term and pull constant factors out. Rewrite a product or quotient as a sum of powers first, for example $\dfrac{x^2+1}{x} = x + \dfrac{1}{x}$. There is no product rule for integrals: $\int x \cdot x^2,dx$ is not $\int x,dx \times \int x^2,dx$.

For a linear inner function, the chain rule backwards gives $$\int f(ax+b),dx = \frac{1}{a}F(ax+b) + c,$$ where $F$ is an antiderivative of $f$. For example $\int \cos(5x),dx = \dfrac{1}{5}\sin 5x + c$ and $\int (2x+1)^5,dx = \dfrac{(2x+1)^6}{12} + c$. The factor $\dfrac{1}{a}$ compensates for the $a$ produced by the chain rule.

Example: Find $\int \left(3x^2 - \dfrac{4}{x} + 2\sqrt{x}\right)dx$. Write $2\sqrt{x} = 2x^{1/2}$ and integrate term by term: $3 \cdot \dfrac{x^3}{3} - 4\ln|x| + 2 \cdot \dfrac{x^{3/2}}{3/2} + c = x^3 - 4\ln|x| + \dfrac{4}{3}x^{3/2} + c$.

Example: Find $\int (2x+1)^5,dx$ and $\int \sin 3x,dx$. For the first, $a = 2$ and $\int u^5,du = \dfrac{u^6}{6}$, so the result is $\dfrac{1}{2}\cdot\dfrac{(2x+1)^6}{6} + c = \dfrac{(2x+1)^6}{12} + c$. For the second, $\int \sin 3x,dx = -\dfrac{1}{3}\cos 3x + c$. Differentiating $\dfrac{(2x+1)^6}{12}$ gives $\dfrac{6 \cdot 2(2x+1)^5}{12} = (2x+1)^5$, which confirms the first.

Finding the constant

If extra information is given, $c$ can be fixed. Given $f'(x)$ and one value of $f$, integrate to get $f(x) + c$ and substitute the known point.

Example: A curve has gradient $\dfrac{dy}{dx} = 6x - 2$ and passes through $(1, 3)$. Find its equation. Integrating, $y = 3x^2 - 2x + c$. At $(1,3)$: $3 = 3 - 2 + c$, so $c = 2$. The curve is $y = 3x^2 - 2x + 2$.

Exam tip: In Paper I, "find $\int \dfrac{(x+1)^2}{x},dx$"-type questions reward expanding first: do not try to integrate a quotient or a product directly. Always write $+c$ for an indefinite integral (it is a mark), and write the modulus in $\ln|x|$. Check by differentiating if time allows. The same skill is needed in every later technique, so slips with $\dfrac{1}{a}$ in $f(ax+b)$ cost marks throughout the paper.