Indefinite Integration – A Level Maths Revision

Indefinite Integration

Integration, a type of calculus, is one of the major new topics at A Level. This revision blog on Indefinite Integration should help you whizz through the topic like there’s no tomorrow (disclaimer: there is a tomorrow….so keep revising)!

The information and exercises can also be downloaded in PDF and PowerPoint format here.

And, if you’d like to practise some of the prior knowledge required to understand this topic, try this multiple-choice quiz, or, for more practice, this choose your own adventure.

If you think you’ve mastered Integration in general, try these exam-style questions.

There are two ways of understanding integration. One is as a way of summing the areas of increasingly small slices to find the area of a whole, particularly where that whole is defined by a non-linear function. To learn more about this, see our revision blog on finding the area under a curve (in progress).

Integration can also be defined as the inverse of differentiation: if you differentiate a function, you get the derivative; if you integrate the derivative you can get back to your original function, with some limitations.

This revision blog will mainly be working with integration as the inverse to differentiation. If you’re not sure about differentiation, catch up with differentiating functions.

When differentiating a power term, you use the following rule:

\frac{d}{dx} x^n = nx^{n-1}

This can be rearranged to say “what power term was differentiated to give π‘₯𝑛?”. Start by increasing each 𝑛 by 1:

\frac{d}{dx} x^{n+1} = (n+1)x^{n}

Then, divide by (𝑛 + 1) (where 𝑛 β‰  -1):


As \frac{1}{n+1} is a constant you can take it inside the derivative without affecting the result:


This gives the general form of the inverse derivative of a power term. Essentially, to find the term that differentiated to π‘₯𝑛, add one to the index (𝑛) and then divide by that value.

Example Question 1

Find f(π‘₯) if f'(π‘₯) = 8π‘₯3

Apply the above formula to find the term that was differentiated to give 8π‘₯3:

\begin{aligned} 8x^3&=8\frac{d}{dx}\frac{x^{3+1}}{3+1}\\&=8\frac{d}{dx}\frac{x^4}{4}\\&=\frac{d}{dx}2x^4 \end{aligned}

At this point, however, you cannot conclusively say that f(π‘₯) = 2π‘₯4. Consider the following three functions and their derivatives:

\begin{aligned} f(x)&=2x^4\\f(x)&=2x^4+9\\f(x)&=2x^4-17\end{aligned}

Each function is different yet each function will have the same derivative. You can address this by saying, if f'(π‘₯) = 8π‘₯3, then:

f(x) = 2x^4 + C

C is a constant added to the function to represent the fact that, during the process of differentiation, any constant terms in the original function are lost.

The process of inverse differentiation is called integration, and is denoted by the long s symbol used below. According to the Fundamental Theorem of Calculus, any continuous function must be integrable.

Integration is the reverse process to differentiation. Any continuous function is integrable over its entire domain.

The power rule of integration says that the indefinite integral of π‘₯𝑛 is given by:

\int{x^n} \,dx =\frac{x^{n+1}}{n+1} + C  for 𝑛 β‰  -1

C is called the β€œconstant of integration”.

This type of integration is called “indefinite” integration because it returns a function, as opposed to “definite” integration which returns a value. The 𝑑π‘₯ at the end of the integral tells you that you are integrating with respect to π‘₯, not any other variables that may be present.

Note that, since integration is the reverse process to differentiation, you can also apply some of the same rules for differentiation to manipulate integrals. Specifically:

The integral of the sum or difference of two or more functions is equal to the sum or difference of the integrals of each function.

\int{f(x) + g(x) \,dx}=\int{f(x)} \,dx + \int{g(x) dx} 

A constant factor can be taken outside of the integral.


Example Question 2


The power rule for integration says that, to integrate a power term, add one to the power and then divide by that new value. You can split this integral up and integrate term by term, taking any constant factors outside of the integral.

\int{x^{\frac{1}{2}}+\frac{3}{x^2}\,dx} = \int{x^{\frac{1}{2}}\,dx} + 3\int{\frac{1}{x^2}\,dx}

Next, rewrite the second integral so it is in the form π‘₯𝑛.

\begin{aligned} &=\int{x^{\frac{1}{2}}\,dx} + 3\int{x^{-2}\,dx} \\ &= \frac{x^\frac{3}{2}}{\frac{3}{2}}+3\times\frac{x^{-1}}{-1}+C\\&=\frac{2}{3}x^{\frac{3}{2}}-\frac{3}{x}+C \end{aligned}

Integrals of this form, which include a constant of integration, are sometimes called the general solution. When you know one or more boundary conditions, you can also find a particular solution.

Example Question 3

Given that 𝑦 = f(π‘₯) passes through (1, 3) and f'(π‘₯) = 9π‘₯2 + 8π‘₯ – 2, find f(π‘₯).

To find the function f(π‘₯), integrate its derivative with respect to π‘₯.

\begin{aligned} f(x) &= \int{9x^2+8x-2}\,dx \\ &= \int{9x^2}\,dx + \int{8x} \, dx + \int{-2} \, dx \\ &= 9\int{x^2}\,dx + 8\int{x} \, dx + \int{-2} \, dx \end{aligned}

By the power rule for integration, the integral of a constant π‘Ž is simply π‘Žπ‘₯, so the integral of -2 becomes -2π‘₯:

\begin{aligned} f(x) &= 9 \times \frac{x^3}{3} + 8 \times \frac{x^2}{2}+(-2x)+C \\ &= 3x^3 + 4x^2-2x + C \end{aligned}

Since f(π‘₯) passes through (1, 3), you can substitute these coordinates into the equation to find the value of C:

\begin{aligned} &f(1) = 3\times 1^3 + 4 \times 1^2 - 2 \times 1 + C = 3 \\ &3 + 4 - 2 + C = 3 \\ &C=-2 \end{aligned}

f(x) = 3x^3 + 4x^2 - 2x - 2

Differentiation can be used to model a variety of problems involving the rate of change of various functions. For instance, consider a function s(𝑑) that describes the displacement of a particle at time t seconds, where 𝑑 β‰₯ 0. By differentiating this function with respect to 𝑑, we can find a function that describes velocity, v(𝑑).

In a similar way, we can reverse this process. Given a specific function for velocity, v(𝑑), we can find a function for displacement, s(𝑑), by integration.

Indefinite Integration – Practice Questions

1. Find the following integrals:

a. \int{x^6}\, dx

b. \int{4x + 2}\, dx

c. \int{x^3-2x^2}\, dx

d. \int{3x^{\frac{1}{3}}}\, dx

e. \int{2x^{\frac{5}{2}}}\, dx

f. \int{6x^{-3}+5x}\, dx

2. Find the following integrals:

a. \int{\frac{4x^3-2x}{x}}\, dx

b. \int{3x(x-5)}\, dx

c. \int{\frac{x^10+x-4}{2x^3}}\, dx

d. \int{(x+5)(3x-1)}\, dx

3. Given that y = f(π‘₯) passes through (2, 1) and f’(π‘₯) = 3π‘₯2 – 5π‘₯, find f(π‘₯).

1. Find the following integrals:

a. \int{x^6}\, dx

\boldsymbol{\frac{x^7}{7}+C }

b. \int{4x + 2}\, dx


c. \int{x^3-2x^2}\, dx


d. \int{3x^{\frac{1}{3}}}\, dx

\boldsymbol{\frac{3x^{\frac{4}{3}}}{\frac{4}{3}}+C=\frac{9x^{\frac{4}{3}}}{4}+C }

e. \int{2x^{\frac{5}{2}}}\, dx

\boldsymbol{ \frac{2x^{\frac{7}{2}}}{\frac{7}{2}}+C=\frac{4x^{\frac{7}{2}}}{7}+C }

f. \int{6x^{-3}+5x}\, dx

\boldsymbol{ \frac{6x^{-2}}{-2}+\frac{5x^2}{2}+C=-\frac{3}{x^2}+\frac{5x^2}{2}+C }

2. Find the following integrals:

a. \int{\frac{4x^3-2x}{x}}\, dx

\boldsymbol{ \int{4x^2-2}\,dx=\frac{4x^3}{3}-2x+C }

b. \int{3x(x-5)}\, dx

\boldsymbol{ \begin{aligned}\int{3x^2-15x}\,dx&=\frac{3x^3}{3}-\frac{15}{2}x^2 + C \\&=x^3-\frac{15}{2}x^2 + C \end{aligned}}

c. \int{\frac{x^{10}+x-4}{2x^3}}\, dx

\boldsymbol{ \begin{aligned} \int{\frac{1}{2}x^7+\frac{1}{2}x^{-2}-2x^{-3}}\,dx&=\frac{1}{16}x^8-\frac{1}{2}x^{-1}+x^{-2}+C\\&=\frac{1}{16}x^8-\frac{1}{2x}+\frac{1}{x^2}+C \end{aligned}}

d. \int{(x+5)(3x-1)}\, dx

\boldsymbol{ \int{3x^2+14x-5}\,dx=x^3+7x^2-5x+C }

3. Given that y = f(π‘₯) passes through (2, 1) and f’(π‘₯) = 3π‘₯2 – 5π‘₯, find f(π‘₯).

\boldsymbol{\begin{aligned} &f(x)=x^3-\frac{5}{2}x^2+C\\&f(2)=-2+C\\&1=-2+C\\&C=3\\&f(x)=x^3-\frac{5}{2}x^2+3 \end{aligned}}

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