1.8.3 Integrating standard functions (A-level only) - Revision Guide

What you'll learn

How integration reverses differentiation.
How to integrate powers, including sums, differences and constant multiples.
How to integrate exponential, reciprocal and trigonometric functions.
How to include the constant of integration and check answers by differentiating.

Integration as reverse differentiation

An antiderivative of a function $f (x)$ is a function whose derivative is $f (x)$ . For example, because

ddx(x3)=3x2,\frac{d}{dx}\left(x^3\right)=3x^2,

an antiderivative of $3x^2$ is $x^3$ .

The symbol for an indefinite integral is

dx.\int f(x)\,dx.

Here, $∫\int$ means “integrate”, $f (x)$ is the integrand being integrated, and $d x$ shows that the variable of integration is $x$ .

Definition

The constant of integration

An indefinite integral represents a family of antiderivatives:

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

where $F^{'} (x) = f (x)$ and $C$ is an arbitrary constant.

The constant is necessary because every constant differentiates to zero. For example, $x^3+2$ , $x^3-7$ and $x^3+100$ all have derivative $3x^2$ .

Common Mistake

Forgetting the constant

Always include $+ C$ when finding an indefinite integral. You do not need a separate constant for every term: one constant at the end represents them all.

Integrating powers of $x$

You already know the differentiation rule

ddx(xn+1)=(n+1)xn.\frac{d}{dx}\left(x^{n+1}\right)=(n+1)x^n.

To reverse this, first increase the power by one, then divide by the new power.

Key Idea

The power rule for integration

For $n≠−1n\neq -1$ ,

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

The restriction $n≠−1n\neq -1$ matters because substituting $n = - 1$ would require division by zero. The function $x−1=1xx^{-1}=\frac{1}{x}$ has a separate integration rule.

Example

Integrating a power

Find $dx\int x^5\,dx$ .

Increase the power by one: $5 + 1 = 6$ .
Divide by the new power:
$dx=x66+C.\int x^5\,dx=\frac{x^6}{6}+C.$
Check by differentiating:
$ddx(x66+C)=x5.\frac{d}{dx}\left(\frac{x^6}{6}+C\right)=x^5.$

The rule also applies to negative and fractional powers, provided the power is not negative one.

Example

Integrating a fractional power

Find $dx\int \sqrt{x}\,dx$ .

Rewrite the root as a power: $x=x1/2\sqrt{x}=x^{1/2}$ .
Increase the power by one:
$12+1=32.\frac{1}{2}+1=\frac{3}{2}.$
Divide by the new power:
$dx=x3/23/2+C=23x3/2+C.\int x^{1/2}\,dx =\frac{x^{3/2}}{3/2}+C =\frac{2}{3}x^{3/2}+C.$

Tip

Divide by a fraction

Dividing by a fraction means multiplying by its reciprocal. For example, dividing by $32\frac{3}{2}$ is the same as multiplying by $23\frac{2}{3}$ .

Sums, differences and constant multiples

Integration is linear. This means you can integrate each term separately, keeping its sign and any constant multiplier:

dx,\int \left(af(x)+bg(x)\right)\,dx =a\int f(x)\,dx+b\int g(x)\,dx,

where $a$ and $b$ are constants.

Example

Integrating a polynomial

Find

dx.\int \left(6x^4-3x^2+8\right)\,dx.

Integrate each term separately:
$dx.\int 6x^4\,dx-\int 3x^2\,dx+\int 8\,dx.$
Apply the power rule, remembering that the constant term is $8x^0$ :
$6(x55)−3(x33)+8x+C.6\left(\frac{x^5}{5}\right) -3\left(\frac{x^3}{3}\right) +8x+C.$
Simplify:
$dx=65x5−x3+8x+C.\int \left(6x^4-3x^2+8\right)\,dx =\frac{6}{5}x^5-x^3+8x+C.$

Common Mistake

Changing a coefficient instead of the power

The integration power rule changes the exponent and divides by the new exponent. Do not simply divide by the original power.

Integrating exponential functions

The function $e^x$ is special because it is unchanged by differentiation:

ddx(ex)=ex.\frac{d}{dx}\left(e^x\right)=e^x.

Therefore,

dx=ex+C.\int e^x\,dx=e^x+C.

For $e^{kx}$ , where $k$ is a non-zero constant, differentiation produces an extra factor of $k$ :

ddx(ekx)=kekx.\frac{d}{dx}\left(e^{kx}\right)=ke^{kx}.

Integration must reverse this multiplication by dividing by $k$ .

Key Idea

Integrating an exponential

For a non-zero constant $k$ ,

dx=1kekx+C.\int e^{kx}\,dx=\frac{1}{k}e^{kx}+C.

Example

Integrating an exponential term

Find $dx\int 7e^{3x}\,dx$ .

The exponent is $3 x$ , so integration introduces a factor of $13\frac{1}{3}$ .
Keep the constant multiplier 7:
$dx=7(13e3x)+C=73e3x+C.\int 7e^{3x}\,dx =7\left(\frac{1}{3}e^{3x}\right)+C =\frac{7}{3}e^{3x}+C.$
Differentiate to check:
$ddx(73e3x)=73⋅3e3x=7e3x.\frac{d}{dx}\left(\frac{7}{3}e^{3x}\right) =\frac{7}{3}\cdot 3e^{3x} =7e^{3x}.$

Integrating $1x\frac{1}{x}$

The power rule cannot integrate $1x\frac{1}{x}$ because $1x=x−1\frac{1}{x}=x^{-1}$ and increasing the power would give zero.

Instead, use the fact that

ddx(ln⁡∣x∣)=1x.\frac{d}{dx}\left(\ln |x|\right)=\frac{1}{x}.

Key Idea

The reciprocal rule

For $x≠0x\neq 0$ ,

dx=ln⁡∣x∣+C.\int \frac{1}{x}\,dx=\ln |x|+C.

The modulus signs allow the formula to apply when $x$ is positive or negative.

Example

Integrating a reciprocal term

Find $dx\int \frac{5}{x}\,dx$ .

Take the constant multiplier outside the integral:
$dx.\int \frac{5}{x}\,dx =5\int \frac{1}{x}\,dx.$
Apply the reciprocal rule:
$dx=5ln⁡∣x∣+C.5\int \frac{1}{x}\,dx =5\ln |x|+C.$

Common Mistake

Do not use the power rule for $x^{- 1}$

The expression $x00\frac{x^0}{0}$ is undefined. Whenever you see an $x^{-1}$ term, use the logarithmic result instead.

Integrating sine and cosine

The standard derivatives are

ddx(sin⁡x)=cos⁡xandddx(cos⁡x)=−sin⁡x.\frac{d}{dx}(\sin x)=\cos x \qquad\text{and}\qquad \frac{d}{dx}(\cos x)=-\sin x.

Reversing these gives

dx=sin⁡x+C\int \cos x\,dx=\sin x+C

and

dx=−cos⁡x+C.\int \sin x\,dx=-\cos x+C.

The negative sign in the second result is essential.

When the angle is $k x$ , differentiation introduces a factor of $k$ . As with $e^{kx}$ , integration reverses this by dividing by $k$ .

Key Idea

Integrating trigonometric functions

For a non-zero constant $k$ ,

dx=1ksin⁡(kx)+C\int \cos(kx)\,dx=\frac{1}{k}\sin(kx)+C

and

dx=−1kcos⁡(kx)+C.\int \sin(kx)\,dx=-\frac{1}{k}\cos(kx)+C.

Example

Integrating a trigonometric expression

Find

dx.\int \left(4\cos(2x)-3\sin(5x)\right)\,dx.

Integrate $4cos⁡(2x)4\cos(2x)$ by dividing by the coefficient 2 inside the angle:
$dx=4(12sin⁡(2x))=2sin⁡(2x).\int 4\cos(2x)\,dx =4\left(\frac{1}{2}\sin(2x)\right) =2\sin(2x).$
Since $dx=−15cos⁡(5x)\int\sin(5x)\,dx=-\frac{1}{5}\cos(5x)$ , account for the existing negative coefficient:
$dx=−3(−15cos⁡(5x))=35cos⁡(5x).\int -3\sin(5x)\,dx =-3\left(-\frac{1}{5}\cos(5x)\right) =\frac{3}{5}\cos(5x).$
Combine the results and add one constant:
$dx=2sin⁡(2x)+35cos⁡(5x)+C.\int \left(4\cos(2x)-3\sin(5x)\right)\,dx =2\sin(2x)+\frac{3}{5}\cos(5x)+C.$

Common Mistake

Losing the sine sign

The integral of sine is negative cosine. In an expression such as $−3sin⁡(5x)-3\sin(5x)$ , this creates two negative signs, so the integrated term is positive.

Combining standard functions

A single integrand may contain powers, exponentials, reciprocals and trigonometric terms. Split it into separate terms and apply the appropriate standard rule to each one.

Example

Integrating a mixed expression

Find

dx.\int \left(3x^2+2e^{-x}+\frac{4}{x}-6\cos(3x)\right)\,dx.

Integrate the power and exponential terms:
$dx=−2e−x.\int 3x^2\,dx=x^3, \qquad \int 2e^{-x}\,dx=-2e^{-x}.$
Integrate the reciprocal and cosine terms:
$dx=−2sin⁡(3x).\int \frac{4}{x}\,dx=4\ln|x|, \qquad \int -6\cos(3x)\,dx=-2\sin(3x).$
Combine the results:
$x^3-2e^{-x}+4\ln|x|-2\sin(3x)+C.$

Tip

Check by differentiating

Differentiate your final expression term by term. This quickly detects missing factors, incorrect signs and forgotten powers.

Exam technique

In the exam

Rewrite roots and reciprocal powers in index form before choosing a rule, but recognise $x^{-1}$ as the special logarithmic case.
For $e^{kx}$ , $sin⁡(kx)\sin(kx)$ and $cos⁡(kx)\cos(kx)$ , divide by the coefficient $k$ inside the exponent or angle.
Keep coefficients and signs attached to their terms, add one $+ C$ at the end, and differentiate your answer to check it.

Self review

Check yourself

Can you integrate $5x^4-2x^{-3}+7$ and simplify the coefficients?
Why does the power rule not apply to $1x\frac{1}{x}$ ?
Can you integrate $3e^{2x}+4\sin(3x)-5\cos x$ without losing any factors or signs?

What you'll learn

How integration reverses differentiation.
How to integrate powers, including sums, differences and constant multiples.
How to integrate exponential, reciprocal and trigonometric functions.
How to include the constant of integration and check answers by differentiating.

Integration as reverse differentiation

An antiderivative of a function $f (x)$ is a function whose derivative is $f (x)$ . For example, because

ddx(x3)=3x2,\frac{d}{dx}\left(x^3\right)=3x^2,

an antiderivative of $3x^2$ is $x^3$ .

The symbol for an indefinite integral is

dx.\int f(x)\,dx.

Here, $∫\int$ means “integrate”, $f (x)$ is the integrand being integrated, and $d x$ shows that the variable of integration is $x$ .

Definition

The constant of integration

An indefinite integral represents a family of antiderivatives:

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

where $F^{'} (x) = f (x)$ and $C$ is an arbitrary constant.

The constant is necessary because every constant differentiates to zero. For example, $x^3+2$ , $x^3-7$ and $x^3+100$ all have derivative $3x^2$ .

Common Mistake

Forgetting the constant

Always include $+ C$ when finding an indefinite integral. You do not need a separate constant for every term: one constant at the end represents them all.

Integrating powers of $x$

You already know the differentiation rule

ddx(xn+1)=(n+1)xn.\frac{d}{dx}\left(x^{n+1}\right)=(n+1)x^n.

To reverse this, first increase the power by one, then divide by the new power.

Key Idea

The power rule for integration

For $n≠−1n\neq -1$ ,

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

The restriction $n≠−1n\neq -1$ matters because substituting $n = - 1$ would require division by zero. The function $x−1=1xx^{-1}=\frac{1}{x}$ has a separate integration rule.

Example

Integrating a power

Find $dx\int x^5\,dx$ .

Increase the power by one: $5 + 1 = 6$ .
Divide by the new power:
$dx=x66+C.\int x^5\,dx=\frac{x^6}{6}+C.$
Check by differentiating:
$ddx(x66+C)=x5.\frac{d}{dx}\left(\frac{x^6}{6}+C\right)=x^5.$

The rule also applies to negative and fractional powers, provided the power is not negative one.

Example

Integrating a fractional power

Find $dx\int \sqrt{x}\,dx$ .

Rewrite the root as a power: $x=x1/2\sqrt{x}=x^{1/2}$ .
Increase the power by one:
$12+1=32.\frac{1}{2}+1=\frac{3}{2}.$
Divide by the new power:
$dx=x3/23/2+C=23x3/2+C.\int x^{1/2}\,dx =\frac{x^{3/2}}{3/2}+C =\frac{2}{3}x^{3/2}+C.$

Tip

Divide by a fraction

Dividing by a fraction means multiplying by its reciprocal. For example, dividing by $32\frac{3}{2}$ is the same as multiplying by $23\frac{2}{3}$ .

Sums, differences and constant multiples

Integration is linear. This means you can integrate each term separately, keeping its sign and any constant multiplier:

dx,\int \left(af(x)+bg(x)\right)\,dx =a\int f(x)\,dx+b\int g(x)\,dx,

where $a$ and $b$ are constants.

Example

Integrating a polynomial

Find

dx.\int \left(6x^4-3x^2+8\right)\,dx.

Integrate each term separately:
$dx.\int 6x^4\,dx-\int 3x^2\,dx+\int 8\,dx.$
Apply the power rule, remembering that the constant term is $8x^0$ :
$6(x55)−3(x33)+8x+C.6\left(\frac{x^5}{5}\right) -3\left(\frac{x^3}{3}\right) +8x+C.$
Simplify:
$dx=65x5−x3+8x+C.\int \left(6x^4-3x^2+8\right)\,dx =\frac{6}{5}x^5-x^3+8x+C.$

Common Mistake

Changing a coefficient instead of the power

The integration power rule changes the exponent and divides by the new exponent. Do not simply divide by the original power.

Integrating exponential functions

The function $e^x$ is special because it is unchanged by differentiation:

ddx(ex)=ex.\frac{d}{dx}\left(e^x\right)=e^x.

Therefore,

dx=ex+C.\int e^x\,dx=e^x+C.

For $e^{kx}$ , where $k$ is a non-zero constant, differentiation produces an extra factor of $k$ :

ddx(ekx)=kekx.\frac{d}{dx}\left(e^{kx}\right)=ke^{kx}.

Integration must reverse this multiplication by dividing by $k$ .

Key Idea

Integrating an exponential

For a non-zero constant $k$ ,

dx=1kekx+C.\int e^{kx}\,dx=\frac{1}{k}e^{kx}+C.

Example

Integrating an exponential term

Find $dx\int 7e^{3x}\,dx$ .

The exponent is $3 x$ , so integration introduces a factor of $13\frac{1}{3}$ .
Keep the constant multiplier 7:
$dx=7(13e3x)+C=73e3x+C.\int 7e^{3x}\,dx =7\left(\frac{1}{3}e^{3x}\right)+C =\frac{7}{3}e^{3x}+C.$
Differentiate to check:
$ddx(73e3x)=73⋅3e3x=7e3x.\frac{d}{dx}\left(\frac{7}{3}e^{3x}\right) =\frac{7}{3}\cdot 3e^{3x} =7e^{3x}.$

Integrating $1x\frac{1}{x}$

The power rule cannot integrate $1x\frac{1}{x}$ because $1x=x−1\frac{1}{x}=x^{-1}$ and increasing the power would give zero.

Instead, use the fact that

ddx(ln⁡∣x∣)=1x.\frac{d}{dx}\left(\ln |x|\right)=\frac{1}{x}.

Key Idea

The reciprocal rule

For $x≠0x\neq 0$ ,

dx=ln⁡∣x∣+C.\int \frac{1}{x}\,dx=\ln |x|+C.

The modulus signs allow the formula to apply when $x$ is positive or negative.

Example

Integrating a reciprocal term

Find $dx\int \frac{5}{x}\,dx$ .

Take the constant multiplier outside the integral:
$dx.\int \frac{5}{x}\,dx =5\int \frac{1}{x}\,dx.$
Apply the reciprocal rule:
$dx=5ln⁡∣x∣+C.5\int \frac{1}{x}\,dx =5\ln |x|+C.$

Common Mistake

Do not use the power rule for $x^{- 1}$

The expression $x00\frac{x^0}{0}$ is undefined. Whenever you see an $x^{-1}$ term, use the logarithmic result instead.

Integrating sine and cosine

The standard derivatives are

ddx(sin⁡x)=cos⁡xandddx(cos⁡x)=−sin⁡x.\frac{d}{dx}(\sin x)=\cos x \qquad\text{and}\qquad \frac{d}{dx}(\cos x)=-\sin x.

Reversing these gives

dx=sin⁡x+C\int \cos x\,dx=\sin x+C

and

dx=−cos⁡x+C.\int \sin x\,dx=-\cos x+C.

The negative sign in the second result is essential.

When the angle is $k x$ , differentiation introduces a factor of $k$ . As with $e^{kx}$ , integration reverses this by dividing by $k$ .

Key Idea

Integrating trigonometric functions

For a non-zero constant $k$ ,

dx=1ksin⁡(kx)+C\int \cos(kx)\,dx=\frac{1}{k}\sin(kx)+C

and

dx=−1kcos⁡(kx)+C.\int \sin(kx)\,dx=-\frac{1}{k}\cos(kx)+C.

Example

Integrating a trigonometric expression

Find

dx.\int \left(4\cos(2x)-3\sin(5x)\right)\,dx.

Integrate $4cos⁡(2x)4\cos(2x)$ by dividing by the coefficient 2 inside the angle:
$dx=4(12sin⁡(2x))=2sin⁡(2x).\int 4\cos(2x)\,dx =4\left(\frac{1}{2}\sin(2x)\right) =2\sin(2x).$
Since $dx=−15cos⁡(5x)\int\sin(5x)\,dx=-\frac{1}{5}\cos(5x)$ , account for the existing negative coefficient:
$dx=−3(−15cos⁡(5x))=35cos⁡(5x).\int -3\sin(5x)\,dx =-3\left(-\frac{1}{5}\cos(5x)\right) =\frac{3}{5}\cos(5x).$
Combine the results and add one constant:
$dx=2sin⁡(2x)+35cos⁡(5x)+C.\int \left(4\cos(2x)-3\sin(5x)\right)\,dx =2\sin(2x)+\frac{3}{5}\cos(5x)+C.$

Common Mistake

Losing the sine sign

The integral of sine is negative cosine. In an expression such as $−3sin⁡(5x)-3\sin(5x)$ , this creates two negative signs, so the integrated term is positive.

Combining standard functions

A single integrand may contain powers, exponentials, reciprocals and trigonometric terms. Split it into separate terms and apply the appropriate standard rule to each one.

Example

Integrating a mixed expression

Find

dx.\int \left(3x^2+2e^{-x}+\frac{4}{x}-6\cos(3x)\right)\,dx.

Integrate the power and exponential terms:
$dx=−2e−x.\int 3x^2\,dx=x^3, \qquad \int 2e^{-x}\,dx=-2e^{-x}.$
Integrate the reciprocal and cosine terms:
$dx=−2sin⁡(3x).\int \frac{4}{x}\,dx=4\ln|x|, \qquad \int -6\cos(3x)\,dx=-2\sin(3x).$
Combine the results:
$x^3-2e^{-x}+4\ln|x|-2\sin(3x)+C.$

Tip

Check by differentiating

Differentiate your final expression term by term. This quickly detects missing factors, incorrect signs and forgotten powers.

Exam technique

In the exam

Rewrite roots and reciprocal powers in index form before choosing a rule, but recognise $x^{-1}$ as the special logarithmic case.
For $e^{kx}$ , $sin⁡(kx)\sin(kx)$ and $cos⁡(kx)\cos(kx)$ , divide by the coefficient $k$ inside the exponent or angle.
Keep coefficients and signs attached to their terms, add one $+ C$ at the end, and differentiate your answer to check it.

Self review

Check yourself

Can you integrate $5x^4-2x^{-3}+7$ and simplify the coefficients?
Why does the power rule not apply to $1x\frac{1}{x}$ ?
Can you integrate $3e^{2x}+4\sin(3x)-5\cos x$ without losing any factors or signs?

1.11.2 Integrating standard functions

What you'll learn

Integration as reverse differentiation

Integrating powers of $x$

Sums, differences and constant multiples

Integrating exponential functions

Integrating $1x\frac{1}{x}$

Integrating sine and cosine

Combining standard functions

Review 1.8.3 Integrating standard functions (A-level only) by teaching Genie

1.11.2 Integrating standard functions

What you'll learn

Integration as reverse differentiation

Integrating powers of $x$

Sums, differences and constant multiples

Integrating exponential functions

Integrating $1x\frac{1}{x}$

Integrating sine and cosine

Combining standard functions

Review 1.8.3 Integrating standard functions (A-level only) by teaching Genie

1.11.2 Integrating standard functions

What you'll learn

Integration as reverse differentiation

Integrating powers of xxx

Sums, differences and constant multiples

Integrating exponential functions

Integrating 1x\frac{1}{x}x1​

Integrating sine and cosine

Combining standard functions

Review 1.8.3 Integrating standard functions (A-level only) by teaching Genie

1.8.3 Integrating standard functions (A-level only) Revision Guide

1.11.2 Integrating standard functions

What you'll learn

Integration as reverse differentiation

Integrating powers of xxx

Sums, differences and constant multiples

Integrating exponential functions

Integrating 1x\frac{1}{x}x1​

Integrating sine and cosine

Combining standard functions

Review 1.8.3 Integrating standard functions (A-level only) by teaching Genie

1.8.3 Integrating standard functions (A-level only) Revision Guide

Integrating powers of $x$

Integrating $1x\frac{1}{x}$

Integrating powers of $x$

Integrating $1x\frac{1}{x}$