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Integration using Partial Fractions

Definition

Integration by partial fractions is an integration technique which uses partial fraction decomposition to simplify the integrand. The integrand is written as partial fractions and then evaluated using standard methods.

Worked Examples

Example 1

Find $\begin{align}\int\dfrac{3x+2}{(x+1)(x+2)}\;\mathrm{d}x.\end{align}$

Solution

To simplify the integral, the integrand $\dfrac{3x+2}{(x+1)(x+2)}$ must be split into partial fractions like so:

\[\dfrac{3x+2}{(x+1)(x+2)}=\dfrac{A}{x+1}+\dfrac{B}{x+2}.\]

This can be simplified by multiplying the equation by $(x+1)(x+2)$:

\begin{align} \dfrac{(3x+2)(x+1)(x+2)}{(x+1)(x+2)} &= \dfrac{A(x+1)(x+2)}{x+1}+\dfrac{B(x+1)(x+2)}{x+2} \\ 3x+2 &= A(x+2)+B(x+1). \end{align}

$A$ is found by setting $x=-1$:

\begin{align} 3\cdot(-1)+2 &= A\bigl((-1)+2\bigl)+B\bigl(x+1\bigl) \\ -3+2 &= A\cdot1+B\cdot0 \\ \Rightarrow A &= -1. \end{align}

Similarly, $B$ is found by setting $x=-2$:

\begin{align} 3\cdot(-2) +2 &= A\bigl((-2)+2\bigl)+B\bigl((-2)+1\bigl) \\ -6+2 &= A\cdot0+B\cdot(-1) \\ \Rightarrow B &= 4. \end{align}

Hence,

\[\dfrac{3x+2}{(x+1)(x+2)}=\dfrac{4}{x+2}-\dfrac{1}{x+1},\]

and the integral becomes:

\[\int\dfrac{3x+2}{(x+1)(x+2)}\;\mathrm{d}x=\int\dfrac{4}{x+2}\;\mathrm{d}x-\int\dfrac{1}{x+1}\;\mathrm{d}x.\]

Recall that the derivative of $\ln \left| f(x) \right|$ is $\dfrac{f'(x)}{f(x)}$. Since differentiation is the opposite of integration (by the fundamental theorem of calculus), this implies:

\[\int\dfrac{f'(x)}{f(x)}\;\mathrm{d}x=\ln{\vert f(x)\vert }+\text{ constant}.\]

Now apply this result to a general function $\dfrac{1}{x+k}$, where $k$ is a constant. Note that the derivative of $x+k$ is $1$. Hence,

\[\int\dfrac{1}{x+k}\;\mathrm{d}x=\ln{\vert x+k\vert }+\text{ constant}.\]

By this result,

\begin{align} \int\dfrac{3x+2}{(x+1)(x+2)}\;\mathrm{d}x &= \int\dfrac{4}{x+2}\;\mathrm{d}x-\int\dfrac{1}{x+1}\;\mathrm{d}x \\ &= 4\int\dfrac{1}{x+2}\;\mathrm{d}x-\int\dfrac{1}{x+1}\;\mathrm{d}x \\ &=4\ln{\vert x+2\vert }-\ln{\vert x+1\vert }+\mathrm{C}, \end{align}

Note that the arbitrary constants of integration arising from each partial fraction have been absorbed into one constant $\mathrm{C}$.

Example 2

Find $\displaystyle{ \int\dfrac{1}{(x-1)^2(x+1)}\;\mathrm{d}x }$.

Solution

Note that there is a repeated factor of $x-1$ in the denominator. The partial fraction decomposition is therefore given by:

\[\dfrac{1}{(x-1)^2(x+1)}=\dfrac{A}{x-1}+\dfrac{B}{(x-1)^2}+\dfrac{C}{x+1}.\]

This can be simplified by multiplying both sides of the equation by $(x-1)^2(x+1)$:

\begin{align} \dfrac{(x-1)^2(x+1)}{(x-1)^2(x+1)} &= \dfrac{A(x-1)^2(x+1)}{x-1} \\ &\quad +\dfrac{B(x-1)^2(x+1)}{(x-1)^2} \\ &\quad+\dfrac{C(x-1)^2(x+1)}{x+1}\\ 1 &= A(x-1)(x+1)+B(x+1)+C(x-1)^2. \end{align}

$B$ is found by setting $x=1$:

\begin{align} 1 &= A(1-1)(1+1)+B(1+1)+C(1-1)^2 \\ &=A\cdot0+B\cdot2+C\cdot 0 \\ \\ \Rightarrow B &= \dfrac{1}{2} \end{align}

$C$ is found by setting $x=-1$:

\begin{align} 1 &= A(-1-1)(-1+1)+B(-1+1)+C(-1-1)^2 \\ &=A\cdot0+B\cdot0+4C \\ \Rightarrow C &= \dfrac{1}{4} \end{align}

Substituting these values for $B$ and $C$ into the expression and setting $x=0$ gives:

\begin{align} 1 &= A(0-1)(0+1)+\dfrac{1}{2}(0+1)+\dfrac{1}{4}(0-1)^2 \\ &= -A+\dfrac{1}{2}+\dfrac{1}{4} \\ \Rightarrow A &= -\dfrac{1}{4} \end{align}

Hence,

\[\dfrac{1}{(x-1)^2(x+1)}=-\dfrac{1}{4(x-1)}+\dfrac{1}{2(x-1)^2}+\dfrac{1}{4(x+1)}\]

and the integral becomes

\[\int\dfrac{1}{(x-1)^2(x+1)}\mathrm{d} x =\int-\dfrac{1}{4(x-1)}+\dfrac{1}{2(x-1)^2}+\dfrac{1}{4(x+1)}\mathrm{d} x.\]

This can be evaluated directly, using standard integrals:

\begin{align} \int\dfrac{1}{4(x-1)}&+\dfrac{1}{2(x-1)^2}+\dfrac{1}{4(x+1)}\mathrm{d} x \\ &= -\dfrac{1}{4}\ln{\vert x-1\vert }-\dfrac{1}{2(x-1)}+\dfrac{1}{4}\ln{\vert x+1\vert }+\mathrm{C}. \end{align}

Video Examples

Example 1

Prof. Robin Johnson uses partial fractions to find $\begin{align}\int\dfrac{x}{x^2-1}\;\mathrm{d}x\end{align}$.

Example 2

Prof. Robin Johnson uses partial fractions to find $\begin{align}\int\dfrac{2x+1}{(x+1)(x-2)}\;\mathrm{d}x\end{align}$ and $\begin{align}\int_0^1 \dfrac{2x+1}{(x+1)(x-2)}\;\mathrm{d}x\end{align}$.

See here for the video showing the partial fraction decomposition.

Example 3

Prof. Robin Johnson uses partial fractions to find $\begin{align}\int\dfrac{x+3}{x(x^2-1)}\;\mathrm{d}x\end{align}$ and $\begin{align}\int_2^3 \dfrac{x+3}{x(x^2-1)}\;\mathrm{d}x\end{align}$.

See here for the video showing the partial fraction decomposition.

Example 4

Prof. Robin Johnson uses partial fractions to find $\begin{align}\int_0^1\dfrac{x+2}{(x+3)(x+4)}\;\mathrm{d}x\end{align}$.

Workbook

This workbook produced by HELM is a good revision aid, containing key points for revision and many worked examples.

Test Yourself

Test yourself: Numbas test on integration using partial fractions

External Resources

Whiteboard maths

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