# Advanced Examples On Integration Set-4

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We now consider more irrational algebraic expressions. The forms that are generally encountered  are listed below along with the recommended substitutions:

(a) \begin{align}\int {\frac{1}{{{L_1}\left( x \right)\sqrt {{L_2}\left( x \right)} }}} \,\,dx \qquad :\;\;\;{\text{Substitute}}\,\,{L_2}\left( x \right) = {t^2}\end{align}

(b) \begin{align}\int {\frac{1}{{L\left( x \right)\sqrt {Q\left( x \right)} }}} \,\,dx \qquad :\;\;\;\;{\text{Substitute}}\,\,L\left( x \right) = \frac{1}{t}\end{align}

(c) \begin{align}\int {\frac{1}{{Q\left( x \right)\sqrt {L\left( x \right)} }}} \,\,dx \qquad :\;\;\;{\text{Substitute}}\,\,L\left( x \right) = {t^2}\end{align}

(d) \begin{align}\int {\frac{1}{{{Q_1}\left( x \right)\sqrt {{Q_2}\left( x \right)} }}} \,\,dx \qquad :\;\;\;\;{\text{Substitute}}\,\,\frac{{{Q_2}\left( x \right)}}{{{Q_1}\left( x \right)}} = {t^2}\left( \begin{gathered}{\text{If the expressions}}{Q_1}\left( x \right){\text{ and }}{Q_2}\left( x \right)\\{\text{are}}\,{\text{purely quadratic}},{\text{i}}{\text{.e}},{\text{the coefficients}}\\{\text{of }}x\,{\text{are both are zero}},{\text{the}}\\{\text{substitution}}\;x = \frac{1}{t}{\text{would also do }}\end{gathered} \right)\end{align}

Some examples will make the use of these substitutions more clear

Example – 56

Evaluate the following integrals:

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

(b) \begin{align}\int {\frac{x}{{\left( {x - 3} \right)\sqrt {x + 1} }}} \,\,dx\end{align}

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

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

Solution: Refer to the substitutions mentioned in the preceeding discussion:

(a) Substitute

\begin{align}& \qquad x+2={{t}^{2}} \\ & \Rightarrow \quad dx=2\,tdt \\ & \Rightarrow\quad I=\int{\frac{1}{\left( x-1 \right)\sqrt{x+2}}}\,\,dx \\ &\qquad\quad =\int{\frac{1}{\left( {{t}^{2}}-3 \right)\cdot t}}\cdot \,2\,tdt \\ &\qquad\quad =2\int{\frac{1}{{{t}^{2}}-3}}\,\,dt \\ &\qquad\quad =\frac{1}{\sqrt{3}}\ln \left| \frac{t-\sqrt{3}}{t+\sqrt{3}} \right|+C \\ &\qquad\quad =\frac{1}{\sqrt{3}}\ln \left| \frac{\sqrt{x+2}-\sqrt{3}}{\sqrt{x+2}+\sqrt{3}} \right|+C \end{align}

(b) Substitute

\begin{align} &\qquad x + 1 = {t^2} \\& \Rightarrow\quad dx=2tdt \\& \Rightarrow \quad I=\int{\frac{x}{\left( x-3 \right)\sqrt{x+1}}}dx \\ & \qquad\quad =\int{\frac{\left( {{t}^{2}}-1 \right)}{\left( {{t}^{2}}-4 \right)\cdot t}}2tdt \\ & \qquad\quad=2\int{\frac{{{t}^{2}}-1}{{{t}^{2}}-4}}dt \\& \qquad\quad=2\int{\left\{ 1+\frac{3}{{{t}^{2}}-4} \right\}}dt \\ & \qquad\quad=2t+\frac{3}{2}\ln \left| \frac{t-2}{t+2} \right|+C \\& \qquad\quad =2\sqrt{x+1}+\frac{3}{2}\ln \left| \frac{\sqrt{x+1}-2}{\sqrt{x+1}+2} \right|+C \end{align}

(c) Substitute

\begin{align}&\qquad x + 1 = \frac{1}{t}\\& \Rightarrow \quad dx=-\frac{1}{{{t}^{2}}}dt \\& \Rightarrow \quad I=\int{\frac{1}{\left( x+1 \right)\sqrt{{{x}^{2}}+x+1}}}dx \\ &\qquad\quad=\int{\frac{t}{\sqrt{{{\left( \frac{1}{t}-1 \right)}^{2}}+\left( \frac{1}{t}-1 \right)+1}}}\cdot \left( -\frac{1}{{{t}^{2}}} \right)dt \\&\qquad\quad=-\int{\frac{1}{\sqrt{{{t}^{2}}-t+1}}}dt \\&\qquad\quad=-\int{\frac{1}{\sqrt{{{\left( t-\frac{1}{2} \right)}^{2}}+\frac{3}{4}}}}dt \\ &\qquad\quad=-\ln \left| \left( t-\frac{1}{2} \right)+\sqrt{{{t}^{2}}-t+1} \right|+C \\& \qquad\quad =-\ln \left| \frac{1}{x+1}-\frac{1}{2}+\frac{\sqrt{{{x}^{2}}+x+1}}{x+1} \right|+C \end{align}

(d) Substitute

\begin{align} &\qquad\quad x+1={{t}^{2}} \\&\Rightarrow \quad dx=2tdt \\&\Rightarrow\quad I=\int{\frac{x}{\left( {{x}^{2}}+2x+2 \right)\sqrt{x+1}}}dx \\&\qquad\quad =\int{\frac{{{t}^{2}}-1}{\left( {{t}^{4}}+1 \right)\cdot t}}\cdot 2tdt \\& \qquad\quad=2\int{\frac{{{t}^{2}}-1}{{{t}^{4}}+1}}dt \end{align}

We’ve already seen how to evaluate integrals of this type