# Advanced Examples On Integration Set-1

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Example - 51

Evaluate the following integrals:

(a) \begin{align}\int {\frac{{{x^2} + x + 1}}{{\sqrt {{x^2} + 2x + 3} }}} \,\,dx\end{align}

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

Solution: Let us first find the constants $$\alpha ,\,\beta \,\,{\rm{and}}\,\,\gamma$$ which will be common to both the questions.

\begin{align}& \,\,{{x}^{2}}+x+1=\alpha \left( {{x}^{2}}+2x+3 \right)+\beta \left( 2x+2 \right)+\gamma \\ & \qquad \qquad\quad =\alpha {{x}^{2}}+\left( 2\alpha +2\beta \right)x+3\alpha +2\beta +\gamma \\ & \qquad \Rightarrow\quad \alpha =1\;,\beta =-\frac{1}{2}\;,\gamma =-1\; \\ \end{align}

(a)                        \begin{align}{I_1} = \int {\frac{{{x^2} + x + 1}}{{\sqrt {{x^2} + 2x + 3} }}} \,\,dx\end{align}

\begin{align}&= \int {\frac{{{x^2} + 2x + 3}}{{\sqrt {{x^2} + 2x + 3} }}} \,\,dx - \frac{1}{2}\int {\frac{{2x + 2}}{{\sqrt {{x^2} + 2x + 3} }}} \,\,dx - \int {\frac{1}{{\sqrt {{x^2} + 2x + 3} }}} \,\,dx\\&= \int {\sqrt {{{\left( {x + 1} \right)}^2} + 2} \,dx} \, - \frac{1}{2}\int {\frac{{2x + 2}}{{\sqrt {{x^2} + 2x + 3} }}} \,\,dx\,\, - \int {\frac{1}{{\sqrt {{{\left( {x + 1} \right)}^2} + 2} }}} \,\,dx\\&\,\,\,\,\left( \begin{array}{l}{\text{of the standard }}\\{\text{form}}\left( {26} \right)\end{array} \right) \qquad \left( \begin{array}{l}{\text{use the substituion}}\\{x^2} + 2x + 3 = t\end{array} \right) \qquad \left( \begin{array}{l}{\text{of the standard }}\\{\text{form}}\left( {22} \right)\end{array} \right)\\ &= \frac{1}{2}\left\{ {\left( {x + 1} \right)\sqrt {{x^2} + 2x + 3} + 2\ln \left| {\left( {x + 1} \right) + \sqrt {{x^2} + 2x + 3} } \right|} \right\} - \sqrt {{x^2} + 2x + 3} - \ln \left\{ {\left( {x + 1} \right) + \sqrt {{x^2} + 2x + 3} } \right\} + C\\&= \frac{1}{2}\left( {x + 1} \right)\sqrt {{x^2} + 2x + 3} \,\,\, - \,\,\,\sqrt {{x^2} + 2x + 3} \,\,\,\, + \,\,\,C\\& = \frac{1}{2}\left( {x - 1} \right)\sqrt {{x^2} + 2x + 3} \,\, + \,\,\,C\end{align}

(b) Using the same values for $$\alpha ,\beta \,\,{\rm{and}}\,\,\gamma ,$$ this integral now becomes

\begin{align}&{I_2} = {\int {\left( {{x^2} + 2x + 3} \right)} ^{3/2}}dx - \frac{1}{2}\int {\left( {2x + 2} \right)\sqrt {{x^2} + 2x + 3} } \,\,dx - \int {\sqrt {{x^2} + 2x + 3} } \,\,dx\\&\qquad\qquad\qquad\qquad\quad\quad \left( \begin{array}{l}{\text{Use the substituion}}\\{x^2} + 2x + 3 = t\end{array} \right) \qquad \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( \begin{array}{l}{\text{of the standard}}\\{\text{form}}\left( \;{26} \right)\end{array} \right)\,\,\,\,\,\,\,\end{align}

The last two integrals can be evaluated as indicated.

To evaluate                                          $${I_3} = {\int {\left( {{x^2} + 2x + 3} \right)} ^{3/2}}dx = \int {{{\left( {{{\left( {x + 1} \right)}^2} + 2} \right)}^{3/2}}} dx,$$

we let                                                                 $$x + 1 = \sqrt 2 \tan \theta$$ so that $$dx = \sqrt 2 {\sec ^2}\theta d\theta :$$

\begin{align} & \Rightarrow \quad {{I}_{3}}=\int{{{2}^{3/2}}}{{\sec }^{3}}\theta .\sqrt{2}{{\sec }^{2}}\theta d\theta : \\ & \qquad\quad\;=4\int{{{\sec }^{5}}\theta d\theta } \\ & \; \Rightarrow\quad I=\frac{{{I}_{3}}}{4}=\int{{{\sec }^{3}}\theta .{{\sec }^{2}}d\theta } \\&\qquad\qquad\qquad\qquad\quad \text{Ist}\quad \text{IInd} \\ &\quad\qquad\;={{\sec }^{3}}\theta \tan \theta -3\int{{{\sec }^{3}}\theta }{{\tan }^{2}}\theta d\theta \\&\quad\qquad\; ={{\sec }^{3}}\theta \tan \theta -3\int{{{\sec }^{3}}\theta }\left( {{\sec }^{2}}\theta -1 \right)d\theta \\ &\quad\qquad\;={{\sec }^{3}}\theta \tan \theta -3\int{{{\sec }^{5}}\theta d\theta +3}\int{{{\sec }^{3}}\theta d\theta } \\&\quad\qquad\;={{\sec }^{3}}\theta \tan \theta -3I+3{{I}_{4}} \\ \end{align}

where $${I_4} = \int {{{\sec }^3}\theta d\theta }$$

$$\Rightarrow \quad 4I = {\sec ^3}\theta \tan \theta + 3{I_4}$$

To evaluate I4 , we again use integration by parts:

\begin{align} & \ \quad \quad {{I}_{4}}=\int{\ \sec \ \theta \ {{\sec }^{2}}\ \theta }d\ \theta \\ & \qquad \qquad \quad \text{Ist}\quad \,\,\text{IInd} \\ & \quad \quad \,\,\,\,\,\,\,\,=\sec \theta \tan \theta -\int{\sec \theta {{\tan }^{2}}\theta d\theta } \\ & \quad \quad \,\,\,\,\,\,\,\,=\sec \theta \tan \theta -\int{\sec \theta \left( {{\sec }^{2}}\theta -1 \right)d\theta } \\ & \quad \quad \quad \,\,\,=\sec \theta \tan \theta -{{I}_{4}}+\int{\sec \theta d\theta } \\ & \Rightarrow \quad 2{{I}_{4}}=\sec \theta \tan \theta +\ln \left| \sec \theta +\tan \theta \right| \\ & \Rightarrow \quad \,\,\,\,\,\text{I}=\frac{1}{4}{{\sec }^{3}}\theta \tan \theta +\frac{3}{8}\sec \theta \tan \theta +\frac{3}{8}\ln \left| \sec \theta +\tan \theta \right|+C \\ \end{align}

Using I, I2 can be determined.

This question was included only for illustration purposes and you should not expect in an actual exam a question with this much detailed calculations required.