# Examples On Integration By Expansion Using Partial Fractions Set-2

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Example - 39

Evaluate \begin{align}\int {\frac{{x + 1}}{{x{{\left( {1 + x{e^x}} \right)}^2}}}} \,\,dx\end{align}

Solution: Observe that the derivative of $$x{e^x}is{\rm{ }}\left( {x + {\rm{ }}1} \right){e^x}$$. Thus, we could substitute $$x{e^x}$$ for a new variable t if we multiply the numerator and denominator of the expression above by $${e^x}$$:

$I = \int {\frac{{x + 1}}{{x{{\left( {1 + x{e^x}} \right)}^2}}}dx}$

$= \int {\frac{{\left( {x + 1} \right){e^x}}}{{x{e^x}{{\left( {1 + x{e^x}} \right)}^2}}}dx}$

The substitution $$x{e^x} = t$$ now reduces I to:

$I = \int {\frac{{dt}}{{t{{\left( {1 + t} \right)}^2}}}}$

We can now expand this expression in t using partial fractions:

\begin{align} & \qquad \qquad \,\qquad \qquad \,\,\,\,\frac{1}{t{{\left( 1+t \right)}^{2}}}=\frac{A}{t}+\frac{B}{1+t}+\frac{C}{{{\left( 1+t \right)}^{2}}} \\ & \qquad \qquad \qquad \qquad\qquad\;\; \Rightarrow \,\,\,1=A{{\left( 1+t \right)}^{2}}+B\left( 1+t \right)t+Ct \\ & \qquad \qquad \,\,\,\,\text{Put}\ t=0\,\qquad\Rightarrow \ A=1 \\ & \qquad \qquad \,\,\,\,\text{Put}\ \ t=-1\quad \Rightarrow \;C=-1 \\ & \;\text{Compare the coeff o}f\;{{t}^{2}}\text{ }\Rightarrow\; \text{ }0=A+B \\ & \qquad \qquad \qquad \qquad \qquad\;\;\Rightarrow \;\; B=-1 \\ \end{align}

The partial fraction expansion is

$\frac{1}{t} - \frac{1}{{1 + t}} - \frac{1}{{{{\left( {1 + t} \right)}^2}}}$

Therefore, I is

\begin{align}& I = \ln \left| t \right| - \ln \left| {1 + t} \right| + \frac{1}{{1 + t}} + C\\ &\;\;= \ln \left| {x{e^x}} \right| - \ln \left| {1 + x{e^x}} \right| + \frac{1}{{1 + x{e^x}}} + C\end{align}

Example - 40

Evaluate \begin{align}\int {\frac{{{x^3} + 3x + 2}}{{{{\left( {{x^2} + 1} \right)}^2}\left( {x + 1} \right)}}} \,\,dx\end{align}

Solution: Instead of directly expanding this expression using partial fractions, a first stage manipulation is possible which will reduce this expression to a simpler form:

\begin{align}& \frac{{{x^3} + 3x + 2}}{{{{\left( {{x^2} + 1} \right)}^2}\left( {x + 1} \right)}} \;= \frac{{x\left( {{x^2} + 1} \right) + 2\left( {x + 1} \right)}}{{{{\left( {{x^2} + 1} \right)}^2}\left( {x + 1} \right)}}\,\\&\qquad\qquad\qquad\qquad= \frac{x}{{\left( {x + 1} \right)\left( {{x^2} + 1} \right)}} + \frac{2}{{{{\left( {{x^2} + 1} \right)}^2}}} \\ &\qquad\qquad\qquad\qquad= {E_1}{\rm{ }}{\rm{ }}{\rm{ }}{\rm{ }}{\rm{ }} + {E_2}\end{align}

We first partially-expand $${E_1}$$ :

\begin{align} & \qquad \,\,\,\frac{x}{\left( x+1 \right)\left( {{x}^{2}}+1 \right)}=\frac{A}{x+1}+\frac{Bx+C}{{{x}^{2}}+1} \\ & \qquad \qquad \qquad \,\,\,\Rightarrow x=A\left( {{x}^{2}}+1 \right)+\left( Bx+C \right)\left( x+1 \right) \\ & \qquad \qquad \qquad Put\,x=-1\Rightarrow\quad A=-\frac{1}{2} \\ & \qquad \qquad \qquad Put\,\,x=0\Rightarrow \quad A+C=0 \\ & \qquad \qquad \qquad \qquad \qquad \Rightarrow \quad C=\frac{1}{2} \\ & \text{Compare the coeff of }{{x}^{2}}\Rightarrow \quad A+B=0 \\ & \qquad \qquad \qquad \qquad \qquad \Rightarrow \quad B=\frac{1}{2} \\ \end{align}

The partial - expansion of $${E_1}$$ therefore is:

$\frac{{ - 1/2}}{{x + 1}} + \frac{{\left( {1/2} \right)x + 1/2}}{{{x^2} + 1}}$

The integral of $${E_1}$$ is

\begin{align}& {I_1} = - \frac{1}{2}\int {\frac{1}{{x + 1}}} \,dx + \frac{1}{2}\int {\frac{x}{{{x^2} + 1}}dx + \frac{1}{2}\int {\frac{1}{{{x^2} + 1}}} \,\,dx}\\ &\quad= - \frac{1}{2}\ln \left| {x + 1} \right| + \frac{1}{4}\ln \left( {{x^2} + 1} \right) + \frac{1}{2}{\tan ^{ - 1}}x + C\end{align}

To integrate $${E_2}$$ , we use the substitution $$x = \tan \theta .$$

Thus, $$dx = {\sec ^2}\theta d\theta :$$

\begin{align}& {I_2} = \int {\frac{{{{\sec }^2}\theta d\theta }}{{{{\left( {{{\sec }^2}\theta } \right)}^2}}}}\\& \quad= \int {{{\cos }^2}\theta d\theta } \\& \quad= \frac{1}{2}\int {\left( {1 + \cos 2\theta } \right)} \,d\theta \\ &\quad= \frac{1}{2}\left( {\theta + \frac{{\sin 2\theta }}{2}} \right) + C\\&\quad = \frac{1}{2}{\tan ^{ - 1}}x + \frac{1}{2}\frac{x}{{{x^2} + 1}} + C\end{align}

We have evaluated the integrals of both$${E_1}$$ and $${E_2}$$ and thus the integral of $${E_1}{\rm{ }}{\rm{ }}{\rm{ }}{\rm{ }}{\rm{ }} + {E_2}$$