Integration basics and standard integrals

Integration reverses differentiation. If $F'(x)=f(x)$, then $F$ is an antiderivative of $f$, and

$$\int f(x)\,dx=F(x)+C.$$

The symbol $\int$ means integrate, $f(x)$ is the integrand, and $dx$ identifies $x$ as the variable of integration. The constant $C$ represents every possible constant because differentiation cannot recover a constant that has disappeared.

Prerequisites

You should be able to:

• use negative and fractional indices;
• simplify algebraic fractions and roots;
• differentiate powers, exponentials and basic trigonometric functions;
• substitute values into a function and solve a linear equation.

Review differentiation and gradients and differentiating standard functions if needed.

Antiderivatives are families of functions

Since

$$\frac{d}{dx}(x^3+7)=3x^2 \quad\text{and}\quad \frac{d}{dx}(x^3-20)=3x^2,$$

the integrand $3x^2$ does not have just one antiderivative. It has the family

$$\boxed{\int 3x^2\,dx=x^3+C}.$$

Different values of $C$ translate the graph vertically but do not change its gradient at any $x$.

Worked example 1: checking by differentiation

Find $\int 12x^3\,dx$.

We need a function whose derivative is $12x^3$. Since differentiating $x^4$ produces $4x^3$, use a coefficient of $3$:

$$\int 12x^3\,dx=3x^4+C.$$

Check:

$$\frac{d}{dx}(3x^4+C)=12x^3.$$

This differentiation check is the most reliable way to detect a missing factor or incorrect power.

The reverse power rule

For $n\ne-1$,

$$\boxed{\int x^n\,dx=\frac{x^{n+1}}{n+1}+C}.$$

Increase the power by $1$, then divide by the new power. The restriction $n\ne-1$ is essential because dividing by the new power would otherwise mean dividing by zero.

More generally,

$$\int ax^n\,dx=\frac{a}{n+1}x^{n+1}+C, \qquad n\ne-1.$$

Worked example 2: integrate a polynomial

Find

$$\int(6x^4-8x^3+5x-7)\,dx.$$

Integrate term by term:

$$\begin{aligned} \int(6x^4-8x^3+5x-7)\,dx &=6\left(\frac{x^5}{5}\right) -8\left(\frac{x^4}{4}\right) +5\left(\frac{x^2}{2}\right)-7x+C\\ &=\boxed{\frac65x^5-2x^4+\frac52x^2-7x+C}. \end{aligned}$$

The integral of the constant $-7$ is $-7x$, not $0$. Differentiation sends constants to zero; integration asks which function differentiates to the given constant.

Worked example 3: negative and fractional powers

Find

$$\int\left(\frac{4}{x^3}+3\sqrt{x}-\frac{2}{\sqrt{x}}\right)dx.$$

Rewrite every term as a power:

$$\frac{4}{x^3}+3\sqrt{x}-\frac{2}{\sqrt{x}} =4x^{-3}+3x^{1/2}-2x^{-1/2}.$$

Now apply the reverse power rule:

$$\begin{aligned} \int\left(4x^{-3}+3x^{1/2}-2x^{-1/2}\right)dx &=\frac{4x^{-2}}{-2}+\frac{3x^{3/2}}{3/2}-\frac{2x^{1/2}}{1/2}+C\\ &=\boxed{-\frac{2}{x^2}+2x^{3/2}-4\sqrt{x}+C}. \end{aligned}$$

Notice that increasing $-3$ by $1$ gives $-2$. It does not make the index positive.

Self-check 1

Find

$$\int\left(5x^4-\frac{6}{x^2}+\sqrt[3]{x}\right)dx.$$

Answer

Write the integrand as $5x^4-6x^{-2}+x^{1/3}$:

$$\begin{aligned} \int\left(5x^4-6x^{-2}+x^{1/3}\right)dx &=x^5+6x^{-1}+\frac34x^{4/3}+C\\ &=\boxed{x^5+\frac6x+\frac34x^{4/3}+C}. \end{aligned}$$

Differentiate each term to verify the signs and coefficients.

Linearity of integration

Constants can be taken outside an integral, and sums can be integrated term by term:

$$\int kf(x)\,dx=k\int f(x)\,dx$$

and

$$\int\bigl(f(x)\pm g(x)\bigr)dx =\int f(x)\,dx\pm\int g(x)\,dx.$$

These rules justify the steps used for polynomials. One final $+C$ is enough. Giving every term its own constant is not wrong, but those constants combine into a single arbitrary constant.

Integration does not distribute across products:

$$\int f(x)g(x)\,dx \ne \left(\int f(x)\,dx\right)\left(\int g(x)\,dx\right)$$

in general. Products may require algebra, integration by substitution or integration by parts.

Standard integrals

The following results reverse standard derivatives. Trigonometric angles must be in radians.

$$\begin{aligned} \int x^n\,dx&=\frac{x^{n+1}}{n+1}+C, && n\ne-1,\\ \int \frac1x\,dx&=\ln|x|+C, && x\ne0,\\ \int e^x\,dx&=e^x+C,\\ \int a^x\,dx&=\frac{a^x}{\ln a}+C, && a>0,\ a\ne1,\\ \int \cos x\,dx&=\sin x+C,\\ \int \sin x\,dx&=-\cos x+C,\\ \int \sec^2x\,dx&=\tan x+C. \end{aligned}$$

The negative sign in $\int\sin x\,dx$ is necessary because

$$\frac{d}{dx}(-\cos x)=\sin x.$$

Why $1/x$ is exceptional

The integrand $1/x=x^{-1}$ cannot use the reverse power rule: increasing the power gives $0$, and division by $0$ is undefined. Instead,

$$\frac{d}{dx}\ln|x|=\frac1x$$

on any interval not containing $0$. The absolute value matters because $1/x$ is defined for negative as well as positive $x$. No single antiderivative can cross $x=0$, where the integrand is undefined.

Worked example 4: combine standard forms

Find

$$\int\left(4e^x+\frac3x-5\sin x+2\sec^2x\right)dx.$$

Using the table term by term gives

$$\boxed{4e^x+3\ln|x|+5\cos x+2\tan x+C}.$$

In particular, $-5\sin x$ integrates to $+5\cos x$ because differentiating $5\cos x$ gives $-5\sin x$.

Worked example 5: an exponential with base $2$

Find $\int 7\cdot2^x\,dx$.

Since $\frac{d}{dx}(2^x)=2^x\ln2$, dividing by $\ln2$ compensates for the extra factor:

$$\boxed{\int 7\cdot2^x\,dx=\frac{7\cdot2^x}{\ln2}+C}.$$

Do not treat $2^x$ as if it were $e^x$.

Simple linear inner functions

When a standard function contains $ax+b$, reverse differentiation introduces a factor of $1/a$:

$$\int f(ax+b)\,dx=\frac1aF(ax+b)+C,$$

provided $F'=f$ and $a\ne0$. This is the chain rule in reverse.

Worked example 6: account for the inner derivative

Find

$$\int\left(e^{3x}+4\cos(2x)-\frac{5}{2x-1}\right)dx.$$

For each term, divide by the derivative of its inner linear expression:

$$\begin{aligned} \int e^{3x}\,dx&=\frac13e^{3x},\\ \int4\cos(2x)\,dx&=2\sin(2x),\\ \int-\frac5{2x-1}\,dx&=-\frac52\ln|2x-1|. \end{aligned}$$

Therefore

$$\boxed{\int\left(e^{3x}+4\cos(2x)-\frac{5}{2x-1}\right)dx =\frac13e^{3x}+2\sin(2x)-\frac52\ln|2x-1|+C}.$$

Differentiate the answer. The inner derivatives $3$, $2$ and $2$ cancel the compensating factors.

This shortcut applies directly only when the inner expression is linear. More general composite functions belong in integration by substitution.

Self-check 2

Find

$$\int\left(6\sin(3x)+\frac4{5x+2}\right)dx.$$

Answer

$$\boxed{-2\cos(3x)+\frac45\ln|5x+2|+C}.$$

Check the first sign carefully:

$$\frac{d}{dx}\bigl(-2\cos(3x)\bigr)=6\sin(3x).$$

Recovering a particular function

An equation for a derivative determines a family of functions. One known point, called an initial condition or boundary condition, determines the value of $C$ and selects one member of that family.

Worked example 7: use an initial condition

Given

$$\frac{dy}{dx}=6x^2-4x+3$$

and $y=7$ when $x=1$, find $y$ in terms of $x$.

First integrate:

$$y=2x^3-2x^2+3x+C.$$

Now use $(x,y)=(1,7)$:

$$7=2(1)^3-2(1)^2+3(1)+C=3+C.$$

Hence $C=4$, so

$$\boxed{y=2x^3-2x^2+3x+4}.$$

Two checks are available. Differentiating gives the stated $dy/dx$, and substituting $x=1$ gives $y=7$.

Worked example 8: recover displacement

A particle moves in a straight line with velocity

$$v(t)=6t-4.$$

At $t=2$, its displacement from a fixed origin is $5$ metres. Find its displacement $s(t)$.

Since $v=ds/dt$,

$$s=\int(6t-4)\,dt=3t^2-4t+C.$$

Use $s(2)=5$:

$$5=3(2)^2-4(2)+C=4+C,$$

so $C=1$. Therefore

$$\boxed{s(t)=3t^2-4t+1\text{ metres}}.$$

The constant has a physical meaning: $s(0)=1$, so the particle was $1$ metre from the origin at time $0$.

Self-check 3

Given

$$\frac{dy}{dx}=3e^{2x}-2\sin x$$

and $y=4$ when $x=0$, find $y$.

Answer

Integrate first:

$$y=\frac32e^{2x}+2\cos x+C.$$

At $x=0$, $e^0=1$ and $\cos0=1$, so

$$4=\frac32+2+C,$$

giving $C=\frac12$. Thus

$$\boxed{y=\frac32e^{2x}+2\cos x+\frac12}.$$

Common misconceptions

• Omitting $+C$: an indefinite integral represents a family. Include $+C$ unless a condition has already fixed it.
• Using the power rule on $x^{-1}$: $\int x^{-1}\,dx=\ln|x|+C$, not an expression involving division by zero.
• Dividing by the old power: increase the power first, then divide by the new power.
• Integrating a constant as zero: $\int k\,dx=kx+C$.
• Losing the trigonometric sign: $\int\sin x\,dx=-\cos x+C$.
• Multiplying by an inner derivative: integration reverses the chain rule, so for a linear inner function you divide by its derivative.
• Splitting products: integration is linear over sums, not products.
• Using degrees: standard trigonometric derivative and integral formulae assume radians.

Mixed retrieval check

Try these without looking back.

1. Find $\int(9x^2-4x^{-3}+6)\,dx$.
2. Find $\int\left(5e^{-2x}+3\cos(4x)\right)dx$.
3. Given $f'(x)=4x-3$ and $f(2)=10$, find $f(x)$.
4. Explain why $\int_a^b f(x)\,dx$ does not need an arbitrary constant in its final value.

Answers

$$\boxed{3x^3+2x^{-2}+6x+C}.$$

The derivative of $2x^{-2}$ is $-4x^{-3}$.

$$\boxed{-\frac52e^{-2x}+\frac34\sin(4x)+C}.$$

3. First integrate:

$$f(x)=2x^2-3x+C.$$

Then $10=2(2)^2-3(2)+C=2+C$, so $C=8$ and

$$\boxed{f(x)=2x^2-3x+8}.$$

4. If $F'(x)=f(x)$, then

$$\int_a^b f(x)\,dx=(F(b)+C)-(F(a)+C)=F(b)-F(a).$$

The same constant occurs at both limits and cancels. The next lesson develops this idea fully.

Next steps

• Learn why an integral describes accumulation in integration as a limit.
• Evaluate integrals with limits and distinguish signed area from geometric area in definite integrals and areas.
• Reverse more general chain rules in integration by substitution.
• Apply antiderivatives to motion in calculus in kinematics.