# Advanced Examples On Integration Set-3

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## Integration examples

Example - 54

Evaluate \begin{align}\int {\frac{{dx}}{{{{\left( {x - 1} \right)}^{3/4}}{{\left( {x + 2} \right)}^{5/4}}}}}\end{align}

Solution: Observe the exponents carefully; \begin{align}\frac{3}{4}\end{align} is “ \begin{align}\frac{1}{4} - {\rm{less}}\end{align} ” than 1 while \begin{align}\frac{5}{4}\,\,\,\text{is}\,\,\,''\frac{1}{4}-\text{more}\end{align} '' than 1. Therefore, we can write the denominator as

${\left( {x - 1} \right)^{3/4}}{\left( {x + 2} \right)^{5/4}} = \left( {x - 1} \right)\left( {x + 2} \right){\left( {\frac{{x + 2}}{{x - 1}}} \right)^{1/4}}$

We thus use the substitution

$\frac{{x + 2}}{{x - 1}} = {t^4}$

$\Rightarrow x = \frac{{{t^4} + 2}}{{{t^4} - 1}}$

$and \;dx = \frac{{ - 12{t^3}}}{{{{\left( {{t^4} - 1} \right)}^2}}}dt$

Thus,                                                                                        \begin{align}I = \int {\frac{{dx}}{{\left( {x - 1} \right)\left( {x + 2} \right){{\left( {\frac{{x + 2}}{{x - 1}}} \right)}^{1/4}}}}} \end{align}

\begin{align}&= \int {\frac{{ - 12{t^3}dt}}{{\left( {\frac{3}{{{t^4} - 1}}} \right)\left( {\frac{{3{t^4}}}{{{t^4} - 1}}} \right) \cdot t \cdot {{\left( {{t^4} - 1} \right)}^2}}}} \\& = - \frac{4}{3}\int {\frac{{dt}}{{{t^2}}}} \\ &= \frac{4}{{3t}} + C\\&= \frac{4}{3}{\left( {\frac{{x - 1}}{{x + 2}}} \right)^{1/4}} + C\end{align}

Example - 55

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

Solution: $I = \int {\left( {x - 1} \right)\sqrt {\frac{{1 + x}}{{1 - x}}} } \,\,dx - \int {\sqrt {\frac{{1 + x}}{{1 - x}}} } \,\,dx$

$= - \int {\sqrt {1 - {x^2}} } \,dx - \int {\sqrt {\frac{{1 + x}}{{1 - x}}} } \,\,dx$

The first integral is of the standard form (27) To evaluate the second integral I1 , we can of course use the substitution $$x = \cos 2\theta .$$ However, here we use the substitution described in the preceeding discussion.

\begin{align} &\quad\quad\frac{1+x}{1-x}={{t}^{2}} \\&\Rightarrow \quad x=\frac{{{t}^{2}}-1}{{{t}^{2}}+1} \\&\Rightarrow \quad dx=\frac{4t}{{{\left( {{t}^{2}}+1 \right)}^{2}}}dt \\ &\Rightarrow \quad {{I}_{1}}=\int{t\cdot \frac{4tdt}{{{\left( {{t}^{2}}+1 \right)}^{2}}}} \\&\qquad\quad\; =4\int{\frac{{{t}^{2}}}{{{\left( {{t}^{2}}+1 \right)}^{2}}}dt} \\&\qquad\quad\;=4\left\{ \int{\frac{1}{{{t}^{2}}+1}dt-\int{\frac{1}{{{\left( {{t}^{2}}+1 \right)}^{2}}}dt}} \right\} \\&\qquad\qquad\qquad \left( \begin{gathered} \text{substitute}\;t=\tan \theta \\\Rightarrow dt={{\sec }^{2}}\theta d\theta \end{gathered} \right) \\ &\qquad\quad\;=4\left\{ {{\tan }^{-1}}t-\int{{{\cos }^{2}}\theta d\theta } \right\} \\&\qquad\quad\; =4\left\{ {{\tan }^{-1}}t-\frac{1}{2}{{\tan }^{-1}}t-\frac{1}{4}\left( \frac{2t}{1+{{t}^{2}}} \right) \right\}+C \\&\qquad\quad\;=2{{\tan }^{-1}}t-\frac{2t}{1+{{t}^{2}}}+C \end{align}

This example once more shows that it is not necessary that there must be only one (unique) substitution for a given integral. Multiple possible substitutions exist for many integrals.