Title: Default for 13.5 Integration by Substitution and Using Partial Fractions
Course: HELM Workbooks 2023-2024

The first technique described here involves making a substitution to simplify an integral. We let a new variable equal a complicated part of the function we are trying to integrate. Choosing the correct substitution often requires experience. This skill develops with practice.

Often the technique of partial fractions can be used to write an algebraic fraction as the sum of simpler fractions. On occasions this means that we can then integrate a complicated algebraic fraction. We shall explore this approach in the second half of the section.

Making a substitution

The technique described here involves making a substitution in order to simplify an integral. We let a new variable, \(u\) say, equal a more complicated part of the function we are trying to integrate. The choice of which substitution to make often relies upon experience: don’t worry if at first you cannot see an appropriate substitution. This skill develops with practice. However, it is not simply a matter of changing the variable - care must be taken with the differential form \(\mathrm{d}x\) as we shall see. The technique is illustrated in the following Example.

Example

Find \({\displaystyle\int (3x+5)^6\ \mathrm{d}x}\).

Solution

First look at the function we are trying to integrate: \((3x+5)^6\). It looks quite complicated to integrate. Suppose we introduce a new variable, \(u\), such that \(u=3x+5\). Doing this means that the function we must integrate becomes \(u^6\). Would you not agree that this looks a much simpler function to integrate than \((3x+5)^6\)? There is a slight complication however. The new function of \(u\) must be integrated with respect to \(u\) and not with respect to \(x\). This means that we must take care of the term \(\mathrm{d}x\) correctly.

Long Method

\(u = 3x + 5\) so \(\dfrac{\mathrm{d}u}{\mathrm{d}x} = 3\) or \(\dfrac{\mathrm{d}x}{\mathrm{d}u} = \dfrac{1}{3}\)

Let \(I = {\displaystyle\int(3x+5)^6\ \mathrm{d}x}\)

\(\phantom{111} = {\displaystyle\int u^6\ \mathrm{d}x}\) (substituting for 3x+5)

\(\phantom{111} = {\displaystyle\int u^6\dfrac{\mathrm{d}x}{\mathrm{d}u}\ \mathrm{d}u}\) (to change from \(x\) to \(u\))

\(\phantom{111} = {\displaystyle\int u^6\dfrac{1}{3}\ \mathrm{d}u}\) (substituting for \(\dfrac{\mathrm{d}x}{\mathrm{d}u} = \dfrac{1}{3}\))

\(\phantom{111} = \dfrac{1}{3}{\displaystyle\int u^6\ \mathrm{d}x} = \dfrac{u^7}{21}+c\)

Short Method

\(u = 3x + 5\) so \(\dfrac{\mathrm{d}u}{\mathrm{d}x} = 3\) so \(\ \mathrm{d}x = \dfrac{1}{3}\ \mathrm{d}u\)

Let \(I = {\displaystyle\int(3x+5)^6\ \mathrm{d}x} = {\displaystyle\int u^6\ \mathrm{d}x} = {\displaystyle\int u^6\dfrac{1}{3}\ \mathrm{d}u} = \dfrac{1}{3}{\displaystyle\int u^6\ \mathrm{d}u} = \dfrac{u^7}{21}+c\)

To finish off we must rewrite this answer in terms of the original variable \(x\) and replace \(u\) by \(3x+5\):

\({\displaystyle\int(3x+5)^6\ \mathrm{d}x} = \dfrac{(3x+5)^7}{21} + c\)

In practice the short method is generally used but mathematicians don’t like to separate the ‘\(\mathrm{d} x\)’ from the ‘\(\mathrm{d} u\)’ as in the statement ‘\(\mathrm{d} x =\dfrac{1}{3}\mathrm{d} u\)’ as it is meaningless mathematically (but it works!). In the future we will use the short method, with apologies to the mathematicians!

Q: Exercise

Use suitable substitutions to find

1. \({\displaystyle\int [value]\ \mathrm{d}x}=\)
2. \({\displaystyle\int [value]\ \mathrm{d}t}=\)
Feedback: 1. We need to calculate the integral \( I =\int [value]\ \mathrm{d}x .\) Using \( u = [value]\) we get \(\dfrac{\mathrm{d}u}{\mathrm{d}x} = [value] \Rightarrow \ \mathrm{d}x = \frac{1}{[value]}\ \mathrm{d}u.\) \[\begin{aligned} I = \int [value]\ \mathrm{d}x = \int u^{[value]}\ \mathrm{d}x = \int \dfrac{u^{[value]}}{[value]}\ \mathrm{d}u = [value]. \end{aligned}\] 2. We need to calculate the integral \( I =\int [value]\ \mathrm{d}t .\) Using \( u = [value]\) we get \(\dfrac{\mathrm{d}u}{\mathrm{d}x} = [value] \Rightarrow \ \mathrm{d}t = \frac{1}{[value]}\ \mathrm{d}u.\) \[\begin{aligned} I = \int [value]\ \mathrm{d}t = \int [value] \sin u \, \mathrm{d}t &= \int [value] \sin u \, \frac{1}{[value]}\ \mathrm{d}u, \\ &= \int \dfrac{\sin (u)}{3}, \\ &= [value].\end{aligned}\]
  Hint: (Hint: you need to simplify sin(t^3 + 1))

Substitution and definite integration

If you are dealing with definite integrals (ones with limits of integration) you must be particularly careful when you substitute. Consider the following example.

Example

Find the definite integral \({\displaystyle\int_2^3 t\sin(t^2)\ \mathrm{d}t}\) by making the substitution \(u=t^2\).

Solution

Note that if \(u=t^2\) then \(\dfrac{\mathrm{d}u}{\mathrm{d}t}=2t\) so that \(\mathrm{d}t = \dfrac{du}{2t}\).

We find \({\displaystyle\int_{t=2}^{t=3}t\sin(t^2)\ \mathrm{d}t}\) = \({\displaystyle\int_{t=2}^{t=3}t\sin u \dfrac{\mathrm{d}u}{2t}} = \dfrac{1}{2}{\displaystyle\int_{t=2}^{t=3}\sin u\ \mathrm{d}u}\)

An important point to note is that the limits of integration are limits on the variable \(t\), not \(u\). To emphasise this they have been written explicitly as \(t=2\) and \(t=3\). When we integrate with respect to the variable \(u\), the limits must be written in terms of \(u\). From the substitution \(u=t^2\), note that when \(t=2\) then \(u=4\) and when \(t=3\) then \(u=9\) so the integral becomes

\(\dfrac{1}{2}{\displaystyle\int_{u=4}^{u=9}\sin u\ \mathrm{d}u} = \dfrac{1}{2}\left[\phantom{\dfrac{1}{1}}\!\!\! -\cos u\right]_4^9= \dfrac{1}{2}\left(-\cos 9 +\cos 4\right) = 0.129\) to 3 d.p.

Q: Exercise

Evaluate the following.

1. \({\displaystyle\int_[value]^[value] [value]\ \mathrm{d}x} =\) (give an exact value)
2. \({\displaystyle\int_[value]^[value] [value]\ \mathrm{d}t} =\) (round to \(4\) decimal places)
Feedback: 1. We need to calculate the integral \( I =\int [value]\ \mathrm{d}x .\) Using \( u = [value]\) we get \(\dfrac{du}{dx} = [value] \Rightarrow \ \mathrm{d}x = \frac{1}{[value]}\ \mathrm{d}u.\) \[ \begin{aligned} I = \int [value]\ \mathrm{d}x = \int u^{[value]}\ \mathrm{d}x &= \int u^{[value]} \frac{1}{[value]}\ \mathrm{d}u, \\ &= \int e^u \ \mathrm{d}u, \\ &= [value]. \end{aligned}\] So \(\displaystyle{\int_[value]^[value] [value]\ \mathrm{d}x = \left[ [value] \right]_[value]^[value] = [value]}.\) 2. We need to calculate the integral \( I =\int [value]\ \mathrm{d}t .\) Using \(u = [value]\) we get \(\dfrac{\mathrm{d}u}{\mathrm{d}t} = [value] \Rightarrow \ \mathrm{d}t = \frac{1}{[value]}\ \mathrm{d}u.\) \(\displaystyle{ I = \int [value]\ \mathrm{d}t = \int [value] e^{u} \, \mathrm{d}t = \int [value] e^u \frac{1}{[value]}\ \mathrm{d}u= [value]}.\) So \(\displaystyle{ \int_[value]^[value] [value]\ \mathrm{d}t = \left[ [value] \right]_[value]^[value]= [value]}.\)
  Hint: (Hint: you need to simplify sin(t^3 + 1))

Integrals giving rise to logarithms

Example

Find \({\displaystyle\int \dfrac{3x^2+1}{x^3+x+2}\ \mathrm{d}x}\)

Solution

Let us consider what happens when we make the substitution \(z=x^3+x+2\). Note that \(\dfrac{\mathrm{d}z}{\mathrm{d}x} = 3x^2+1\) so that we can write \(\mathrm{d}z = (3x^2+1)\mathrm{d}x\)

Then \({\displaystyle\int \dfrac{3x^2+1}{x^3+x+2}\ \mathrm{d}x} = {\displaystyle\int \dfrac{1}{z}\ \mathrm{d}z} = \ln |z| + c = \ln |x^3+x+2| + c\)

Note that in the last Example, the numerator of the integrand \((3x^2+1)\) is the derivative of the denominator \((x^3+x+2)\). The result is the logarithm of the denominator. This is a special case of the following rule:

Key Point

\({\displaystyle\int \dfrac{f'(x)}{f(x)}\ \mathrm{d}x} = \ln |f(x)| +c\)

Note that it is the modulus of \(f(x)\) in the answer.

Q: Exercise

Write down the result of finding the following integrals.

1. \({\displaystyle\int [value]\ \mathrm{d}x}=\)
2. \({\displaystyle\int [value]\ \mathrm{d}t}=\)
3. \({\displaystyle\int [value]\ \mathrm{d}x}=\)
4. \({\displaystyle\int [value]\ \mathrm{d}x}=\)
Feedback: 1. \({\displaystyle\int [value]\ \mathrm{d}x}=[value] \int \dfrac{1}{x}\ \mathrm{d}x = [value]\) 2. Using \( z= t^{[value]} +1\), we observe that \({\displaystyle\int [value]\ \mathrm{d}t= \int \frac{z'}{z}\ \mathrm{d}t}.\) Hence, \({\displaystyle\int [value]\ \mathrm{d}t= \ln|z|+c = [value]}.\) 3. Using \( z= [value]\), we observe that the derivative of the denominator is [value] whereas the numerator is just \(1.\) We adjust the numerator as follows: \({\displaystyle{ \int [value]\ \mathrm{d}x}=\frac{1}[value] \int [value]\ \mathrm{d}x = [value]}.\) 4. Using \( z= [value]\), we observe that the derivative of the denominator is [value] whereas the numerator is just \([value].\) We adjust the numerator as follows: \({\displaystyle{ \int [value]\ \mathrm{d}x}=\frac{[value]}[value] \int [value]\ \mathrm{d}x = [value] [value] + c=[value]}.\)
  Hint: (Hint: you need to simplify sin(t^3 + 1))

Integration using partial fractions

Sometimes expressions which at first sight look impossible to integrate using the techniques already met may in fact be integrated by first expressing them as simpler partial fractions, and then using the techniques described earlier in this Section. Consider the following Example.

Example

Express \(\dfrac{23-x}{(x - 5)(x + 4)}\) as the sum of its partial fractions.

Hence find \({\displaystyle\int \dfrac{23-x}{(x - 5)(x + 4)}\ \mathrm{d}x}\)

Solution

First produce the partial fractions. Write the fraction in the form \(\dfrac{A}{x - 5} + \dfrac{B}{x + 4}\) and find \(A\), \(B\) to obtain \(A = 2\) and \(B = -3\)

Now integrate each term separately:

\({\displaystyle\int \dfrac{23-x}{(x - 5)(x + 4)}\ \mathrm{d}x}\)

\(= {\displaystyle\int \dfrac{A}{x - 5}\ \mathrm{d}x} + {\displaystyle\int \dfrac{B}{x + 4}\ \mathrm{d}x}\)

\(= {\displaystyle\int \dfrac{2}{x - 5}\ \mathrm{d}x} - {\displaystyle\int \dfrac{3}{x + 4}\ \mathrm{d}x}\)

\(= 2\ln|x - 5| - 3\ln|x + 4|\)

Q: Exercise

By expressing the following in partial fractions, evaluate each integral:

1. \({\displaystyle\int [value]\ \mathrm{d}x}=\)
2. \({\displaystyle\int [value]\ \mathrm{d}x}=\)
3. \({\displaystyle\int [value]\ \mathrm{d}x}=\)
4. \({\displaystyle\int [value]\ \mathrm{d}x}=\)
Feedback: We express the integral in partial fractions, and evaluate each integral using the substitution \( z = \) denominator. 1. Note that, since we have improper fraction, \(\begin{aligned} [value] &= \dfrac{A}{x+1} + \frac{B}{(x+1)^2} + Cx + D, \\ \equiv x^3 &= A(x+1) + B + Cx(x+1)^2 + D(x+1)^2 \end{aligned}.\) Equating coefficients yields the partial fraction decomposition: \(\displaystyle [value] = \displaystyle[value]\). Therefore \(\begin{aligned} \displaystyle\int [value] \ \mathrm{d}x &= \displaystyle\int [value] \ \mathrm{d}x, \\ &= [value]. \end{aligned}\) 2. Note that \(\displaystyle [value] = \displaystyle [value]\). This is a proper fraction, so \( \displaystyle[value] = \dfrac{A}{2x+1} + \dfrac{B}{3x-2}\). Multiplying both sides by the denominator and equating coefficients yields the partial fraction decomposition: \( \displaystyle [value] = \displaystyle [value]\). Therefore \(\begin{aligned}\displaystyle\int [value]\ \mathrm{d}x &= \displaystyle\int [value]\ \mathrm{d}x, \\ &= [value] \end{aligned}\). 3. Since we have a proper fraction, we have \(\displaystyle [value] = \dfrac{A}{x-5} + \dfrac{B}{x+1}\). Multiplying both sides by the denominator and equating coefficients yields the partial fraction decomposition: \( \displaystyle [value] = \displaystyle [value]\). Therefore \(\begin{aligned}\displaystyle\int [value]\ \mathrm{d}x &= \displaystyle\int [value]\ \mathrm{d}x, \\ &= [value] \end{aligned}\). 4. Since we have a proper fraction, we have \(\displaystyle [value] = \dfrac{A}{x+1} + \dfrac{B}{x-1} + \dfrac{C}{(x-1)^2}\). Multiplying both sides by the denominator and equating coefficients yields the partial fraction decomposition: \( \displaystyle [value] = \displaystyle [value]\). Therefore \(\begin{aligned}\displaystyle\int [value]\ \mathrm{d}x &= \displaystyle\int [value]\ \mathrm{d}x, \\ &= [value] \end{aligned}\).
  Hint: (Hint: you need to simplify sin(t^3 + 1))

Exercise

By making the substitution \(u=\sin x\) find \({\displaystyle\int [value]\ \mathrm{d}x}\).

1. First find \(\dfrac{\mathrm{d}u}{\mathrm{d}x} =\)
2. Now make the substitution, simplify the result and finally perform the integration. \({\displaystyle\int [value]\ \mathrm{d}x} =\)
Feedback: 1. You are given the substitution \(u=\sin x\). Then \(\dfrac{\mathrm{d}u}{\mathrm{d}x} = [value]\). 2. Substitute \(u=\sin x\) and \(\dfrac{\mathrm{d}u}{\mathrm{d}x} = [value]\) into \({\displaystyle\int [value]\ \mathrm{d}x}\). Then \(\begin{aligned}\displaystyle\int [value] \cdot \dfrac{\mathrm{d}u}{\mathrm{d}x} \ \mathrm{d}x &= \displaystyle\int [value]\ \mathrm{d}u, \\ &= [value] + c, \\ &= [value] \end{aligned}\).

Exercise

Write down, purely by inspection, the following integrals:

1. \({\displaystyle\int \dfrac{1}{x + 1}\ \mathrm{d}x} =\)
2. \({\displaystyle\int \dfrac{2x}{x^2 + 8}\ \mathrm{d}x} =\)
3. \({\displaystyle\int \dfrac{1}{x - 3}\ \mathrm{d}x} =\)

Hint: In each case the numerator of the integrand is the derivative of the denominator.
Feedback: For each part we make use of \({\displaystyle\int \dfrac{f'(x)}{f(x)}\ \mathrm{d}x} = \ln |f(x)| +c\). 1. Note that \( f(x) = x+[value] \implies f'(x) = [value]\). Then \[\displaystyle\int [value]\ \mathrm{d}x = [value].\] 2. Note that \( f(x) = x^2+[value] \implies f'(x) = [value]\). Then \[\displaystyle\int [value]\ \mathrm{d}x = [value].\] 3. Note that \( f(x) = x-[value] \implies f'(x) = [value]\). Then \[\displaystyle\int [value]\ \mathrm{d}x = [value].\]

Exercise

Evaluate the definite integral \(\displaystyle\int^4_2 [value] \ \mathrm{d}t =\) .

Note: Please give your answer to 3 decimal places.
Feedback: We make use of the fact that \({\displaystyle\int \dfrac{f'(x)}{f(x)}\ \mathrm{d}x} = \ln |f(x)| +c\). Let \(f(x) = [value]\). Then \(f'(x) = [value]\). Therefore \(\displaystyle\int^4_2 [value] \ \mathrm{d}t = \bigg[[value] \bigg]^4_2 = [value] = [value]\) to 3 d.p.

Exercise

Express \(\displaystyle [value]\) as the sum of its partial fractions and hence find \(\displaystyle\int [value]\ \mathrm{d}x\).

1. First write \(\displaystyle [value]\) in the form \(\dfrac{A}{x-[value]} + \dfrac{B}{x+[value]}\).
\(A =\) and \(B =\)
2. Then \(\displaystyle\int [value]\ \mathrm{d}x =\)
Feedback: 1. \(\displaystyle [value]\) is a proper fraction, therefore the partial fraction decomposition is of the form \(\displaystyle [value] = \dfrac{A}{x-[value]} + \dfrac{B}{x+[value]}\). Multiplying both sides by the denominator and equating coefficients yields \(\displaystyle [value] = \displaystyle [value]\). So \(A = \displaystyle [value]\) and \(B = \displaystyle [value]\). 2. \(\begin{aligned} \displaystyle\int [value] \ \mathrm{d}x &= \displaystyle\int [value]\ \mathrm{d}x, \\ &= [value] \end{aligned}\).

Example

Evaluate the definite integral:

\({\displaystyle\int \dfrac{x^2}{x^3 +1 }\ \mathrm{d}x}\)

Solution

In this Example the derivative of the denominator is \(3x^2\) whereas the numerator is just \(x^2\). We adjust the numerator as follows:

\({\displaystyle\int \dfrac{x^2}{x^3 +1 }\ \mathrm{d}x} = {\displaystyle\dfrac{1}{3}\int \dfrac{3x^2}{x^3 +1 }\ \mathrm{d}x}\)

and integrate by the rule to get

\(\dfrac{1}{3}\ln|x^3 + 1| + c\)

Note that the sort of procedure in the last Example is only possible because we can move constant factors through the integral sign. It would be wrong to try to move terms involving the variable \(x\) in a similar way.

Exercise

Express \(\displaystyle [value]\) as the sum of its partial fractions and hence find \(\displaystyle\int [value]\ \mathrm{d}x\).

1. First write \(\displaystyle [value]\) in the form \(\dfrac{A}{x-[value]} + \dfrac{B}{x+[value]}\).
\(A =\) and \(B =\)
2. Then \(\displaystyle\int [value]\ \mathrm{d}x =\)
Feedback: 1. \(\displaystyle [value]\) is a proper fraction, therefore the partial fraction decomposition is of the form \(\displaystyle [value] = \dfrac{A}{x-[value]} + \dfrac{B}{x+[value]}\). Multiplying both sides by the denominator and equating coefficients yields \(\displaystyle [value] = \displaystyle [value]\). So \(A = \displaystyle [value]\) and \(B = \displaystyle [value]\). 2. \(\begin{aligned} \displaystyle\int [value] \ \mathrm{d}x &= \displaystyle\int [value]\ \mathrm{d}x, \\ &= [value] \end{aligned}\)