Examples On Integration By Expansion Using Partial Fractions Set-2

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 \({E_1}{\rm{ }}{\rm{ }}{\rm{ }}{\rm{ }}{\rm{ }} + {E_2}\)