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INTEGRALS — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsINTEGRALS5 Marks
Question
Evaluate the integral in Exercise:
$\int^{2}_{0}\frac{\text{dx}}{\text{x}+4-\text{x}^{2}}$
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Answer

$\text{Let}\ \text{I}=\int\limits_{0}^{2}\frac{\text{dx}}{\text{x}+4-\text{x}^{2}}=\int\limits_{0}^{2}\frac{\text{dx}}{4-(\text{x}^{2}-\text{x})}$
$=\int\limits_{0}^{2}\frac{\text{dx}}{\text{x}+4-\text{x}^{2}}=\int\limits_{0}^{2}\frac{\text{dx}}{4-(\text{x}^{2}-\text{x})}$
$=\int\limits_{0}^{2}\frac{\text{dx}}{\bigg(4+\frac{1}{4}\bigg)-\bigg(\text{x}^{2}-\text{x}+\frac{1}{4}\bigg)}=\int\limits_{0}^{2}\frac{\text{dx}}{\bigg(\frac{\sqrt{17}}{2}\bigg)^{2}-\bigg(\text{x}-\frac{1}{2}\bigg)^{2}}$
$=\frac{1}{2\times\frac{\sqrt{17}}{2}}\begin{bmatrix} \log\left|\frac{\frac{\sqrt{17}}{2}+\bigg(\text{x}-\frac{1}{2}\bigg)}{\frac{\sqrt{17}}{2}-\bigg(\text{x}-\frac{1}{2}\bigg)}\right| \\ \end{bmatrix}^{2}_{0}$
$=\frac{1}{\sqrt17}\begin{bmatrix} \log\left\{\frac{\frac{\sqrt{17}}{2}+\bigg({2}-\frac{1}{2}\bigg)}{\frac{\sqrt{17}}{2}-\bigg({2}-\frac{1}{2}\bigg)}\right\}-\log\left\{\frac{\frac{\sqrt{17}}{2}-\frac{1}{2}}{\frac{\sqrt{17}}{2}+\frac{1}{2}}\right\} \\ \end{bmatrix}$
$={\frac{\sqrt{1}}{17}}\bigg[\log\frac{\sqrt{17+3}}{\sqrt{17}-3}-\log\frac{\sqrt{17}-1}{\sqrt{17}+1}\bigg]$
$={\frac{\sqrt{1}}{17}}\log\bigg[\frac{\sqrt{17+3}}{\sqrt{17}-3}\times\frac{\sqrt{17}+1}{\sqrt{17}-1}\bigg]=\frac{1}{\sqrt{17}}\log\bigg(\frac{17+3+4\sqrt{17}}{17+3+-\sqrt{17}}\bigg)$
$={\frac{\sqrt{1}}{17}}\log\bigg(\frac{5+\sqrt{17}}{5-\sqrt{17}}\bigg)=\frac{1}{17}\log\bigg(\frac{5+\sqrt{17}}{5-\sqrt{17}}\bigg)=\frac{1}{\sqrt{17}}\log\bigg(\frac{5+\sqrt{17}}{5-\sqrt{17}}\times\frac{5+\sqrt{17}}{5+\sqrt{17}}\bigg)$
$={\frac{\sqrt{1}}{17}}\log\bigg[\frac{25+17+10\sqrt{17}}{25-17}\bigg]=\frac{1}{\sqrt{17}}\log\bigg[\frac{42+10\sqrt{17}}{8}\bigg]$
$={\frac{\sqrt{1}}{17}}\log\bigg[\frac{21+5\sqrt{7}}{4}\bigg]$

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