Why the modulus in |x|?

For negative x, (-x) also differentiates to 1x. The modulus covers both cases in one formula.

1.08PureCore

Indefinite Integrals

Q: Can I integrate (2x + 1)^3 without expanding?

Yes, because the inside is linear: (2x+1)^44 2 + c. For non-linear insides you need substitution or expansion.

Differentiation takes a function and gives its gradient. $\textbf{Integration}$ goes the other way: given the gradient, find the function. Because every constant differentiates to zero, the answer is never unique — there is a whole family of curves with the same gradient, which is why every indefinite integral carries a $+\,c$ .

40 min Integration

The big picture

The Fundamental Theorem of Calculus connects two ideas that look unrelated: undoing differentiation, and finding the area under a curve. This lesson is the first half — reverse differentiation — and every rule in it is simply a differentiation rule read backwards. Where differentiation lowers a power by one, integration raises it; where differentiating $\sin$ gives $\cos$ , integrating $\cos$ gives $\sin$ . The constant of integration is not a formality: it represents a genuinely unknown vertical position, and a single known point on the curve is enough to pin it down. That idea returns in kinematics, where integrating acceleration needs initial conditions, and in differential equations.

What you'll be able to do

Understand integration as the reverse of differentiation
Integrate $x^{n}$ ( $n \neq -1$ ), $\dfrac{1}{x}$ , $\mathrm{e}^{kx}$ , $\sin kx$ , $\cos kx$ and $\sec^{2}kx$
Include and interpret the constant of integration
Find the equation of a curve from its gradient function and a point

Reversing differentiation

If $\dfrac{\mathrm{d}}{\mathrm{d}x}\mathrm{F}(x) = \mathrm{f}(x)$ , then $\int \mathrm{f}(x)\,\mathrm{d}x = \mathrm{F}(x) + c$ . The $\mathrm{d}x$ says the variable is $x$ .

Since $x^{3}$ , $x^{3} + 5$ and $x^{3} - 2$ all differentiate to $3x^{2}$ , the integral of $3x^{2}$ is $x^{3} + c$ for an arbitrary constant $c$ .

You can always check an integral by differentiating your answer.

Why this works

Geometrically, the curves $y = x^{3} + c$ are vertical translations of each other. At every value of $x$ they have the same gradient, so the gradient alone cannot tell them apart.

Powers of x

To integrate $x^{n}$ : raise the power by one and divide by the new power. This works for every rational $n$ except $n = -1$ , which would mean dividing by zero.

Constants multiply through, and sums integrate term by term. Rewrite roots and fractions as powers first, and expand brackets — there is no product rule for integration.

\int x^{n}\,\mathrm{d}x = \dfrac{x^{n+1}}{n+1} + c, \quad n \neq -1

Power rule for integration.

Example 1Find

\int \dfrac{(x + 2)^{2}}{\sqrt{x}}\,\mathrm{d}x

1Expand:

(x + 2)^{2} = x^{2} + 4x + 4

2Divide by

x^{1/2}

x^{3/2} + 4x^{1/2} + 4x^{-1/2}

3Integrate each:

\dfrac{x^{5/2}}{5/2} + \dfrac{4x^{3/2}}{3/2} + \dfrac{4x^{1/2}}{1/2}

Answer

\tfrac25 x^{5/2} + \tfrac83 x^{3/2} + 8x^{1/2} + c

Tip — Dividing by a fraction like $\tfrac52$ is multiplying by $\tfrac25$ . Write the reciprocal straight away rather than leaving a compound fraction.

Exponentials, logarithms and trigonometric functions

The missing case $n = -1$ is supplied by the natural log: $\int \dfrac{1}{x}\,\mathrm{d}x = \ln|x| + c$ . The modulus allows negative $x$ .

$\mathrm{e}^{kx}$ integrates to $\dfrac{1}{k}\mathrm{e}^{kx}$ — divide by $k$ , the reverse of multiplying by $k$ when differentiating.

In radians: $\int \cos kx\,\mathrm{d}x = \dfrac{1}{k}\sin kx + c$ , $\int \sin kx\,\mathrm{d}x = -\dfrac{1}{k}\cos kx + c$ , and $\int \sec^{2}kx\,\mathrm{d}x = \dfrac{1}{k}\tan kx + c$ .

For a linear inside $(ax + b)$ , the same results hold with a factor of $\dfrac{1}{a}$ : $\int (3x - 1)^{5}\,\mathrm{d}x = \dfrac{(3x - 1)^{6}}{18} + c$ .

\int \dfrac{1}{x}\,\mathrm{d}x = \ln|x| + c

The n = -1 case.

\int \mathrm{e}^{kx}\,\mathrm{d}x = \dfrac{1}{k}\mathrm{e}^{kx} + c

Exponential.

\int \sin kx\,\mathrm{d}x = -\dfrac{1}{k}\cos kx + c

Note the minus sign.

Example 1Find

\int \left(4\mathrm{e}^{2x} - \dfrac{3}{x} + 6\sin 3x\right)\mathrm{d}x

\int 4\mathrm{e}^{2x}\,\mathrm{d}x = 2\mathrm{e}^{2x}

\int \dfrac{3}{x}\,\mathrm{d}x = 3\ln|x|

\int 6\sin 3x\,\mathrm{d}x = -2\cos 3x

Answer

2\mathrm{e}^{2x} - 3\ln|x| - 2\cos 3x + c

Finding the constant

If you know the gradient function and one point on the curve, integrate, then substitute the point to find $c$ .

The same method applies whenever a rate of change and an initial value are given — velocity from acceleration, volume from flow rate.

Example 1A curve has

\dfrac{\mathrm{d}y}{\mathrm{d}x} = 6x^{2} - 4x + 1

and passes through

(2, 9)

. Find its equation.

1Integrate:

y = 2x^{3} - 2x^{2} + x + c

2Substitute

(2, 9)

9 = 16 - 8 + 2 + c

, so

c = -1

Answer

y = 2x^{3} - 2x^{2} + x - 1

Tip — Losing the $+\,c$ costs a mark on every indefinite integral, and makes "find the curve" questions impossible. Write it on the same line as the integration.

Think like an examiner

Include $+\,c$ in every indefinite integral.
Rewrite roots and reciprocals as powers, and expand products, before integrating.
Use $\ln|x|$ for $\int \tfrac1x\,\mathrm{d}x$ .
Check by differentiating your answer.

Common misconceptions

Integrating

x^{-1}

\dfrac{x^{0}}{0}

\int x^{-1}\,\mathrm{d}x = \ln|x| + c

Integrating

\sin x

\cos x

\int \sin x\,\mathrm{d}x = -\cos x + c

Integrating

\mathrm{e}^{4x}

4\mathrm{e}^{4x}

Divide by 4:

\tfrac14\mathrm{e}^{4x} + c

Integrating a product factor by factor.

Expand first; there is no simple product rule for integrals.

Standard integrals

\int x^{n}\,\mathrm{d}x = \tfrac{x^{n+1}}{n+1} + c

n \neq -1

\int \tfrac1x\,\mathrm{d}x = \ln|x| + c

Reciprocal.

\int \mathrm{e}^{kx}\,\mathrm{d}x = \tfrac1k\mathrm{e}^{kx} + c

Exponential.

\int \cos kx\,\mathrm{d}x = \tfrac1k\sin kx + c

Cosine.

Stretch yourself

A curve passes through $\left(\tfrac{\pi}{2}, 3\right)$ and has gradient function $\dfrac{\mathrm{d}y}{\mathrm{d}x} = 2\cos 2x + \sin x$ . Find the equation of the curve and its $y$ -intercept.

Hint — Integrate each term with its $\tfrac1k$ factor, remembering the sign for sine. Then substitute $x = \tfrac{\pi}{2}$ .

Reveal solution

SolutionIntegrating:

y = \sin 2x - \cos x + c

. At

x = \tfrac{\pi}{2}

\sin\pi - \cos\tfrac{\pi}{2} + c = 0 - 0 + c = 3

, so

c = 3

. The curve is

y = \sin 2x - \cos x + 3

. At

x = 0

y = 0 - 1 + 3 = 2

, so the

y

-intercept is

(0, 2)

. Check by differentiating:

2\cos 2x + \sin x

✓.

\blacksquare

Questions students ask

Why is there a

+\,c

Constants differentiate to zero, so reversing differentiation cannot tell you what constant was there.

c

stands for every possibility.

Why the modulus in

\ln|x|

For negative

x

\ln(-x)

also differentiates to

\tfrac1x

. The modulus covers both cases in one formula.

Can I integrate

(2x + 1)^{3}

without expanding?

Yes, because the inside is linear:

\dfrac{(2x+1)^{4}}{4 \times 2} + c

. For non-linear insides you need substitution or expansion.

How do I check an integral?

Differentiate your answer. You should get back exactly the function you integrated.

Key takeaways

Integration reverses differentiation; indefinite integrals include $+\,c$ .
$\int x^{n}\,\mathrm{d}x = \tfrac{x^{n+1}}{n+1} + c$ for $n \neq -1$ ; $\int \tfrac1x\,\mathrm{d}x = \ln|x| + c$ .
$\int \mathrm{e}^{kx}\,\mathrm{d}x = \tfrac1k\mathrm{e}^{kx} + c$ ; trig integrals need radians and care with signs.
Rewrite as powers and expand before integrating.
A known point determines the constant.