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Definite Integrals — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsDefinite Integrals4 Marks
Question
Evaluate the following definite integrals:
$\int\limits_{0}^{2}\frac{1}{4+\text{x}-\text{x}^2}\text{ dx}$
✓

Answer

Let $\text{I}=\int_{0}^\limits{2}\frac{1}{4+\text{x}-\text{x}^2}\text{ dx}$ Then
$\text{I}=\int_{0}^\limits{2}\frac{1}{\text{x}^2-\text{x}-4}\text{ dx}$
$\Rightarrow\text{I}=-\int_{0}^\limits{2}\frac{1}{\Big(\text{x}^2-\text{x}+\frac{1}{4}\Big)-\frac{1}{4}-4}\text{ dx}$
$\Rightarrow\text{I}=-\int_{0}^\limits{2}\frac{1}{\Big(\text{x}-\frac{1}{2}\Big)^2-\frac{17}{4}}\text{ dx}$
$\Rightarrow\text{I}=-\int_{0}^\limits{2}\frac{1}{\Big(\text{x}-\frac{1}{2}\Big)^2-\Big(\frac{\sqrt{17}}{2}\Big)^2}\text{ dx}$
$\Rightarrow\text{I}=\int_{0}^\limits{2}\frac{1}{-\big(\frac{2\text{x}-1}{2}\big)^2+\big(\frac{\sqrt{17}}{2}\big)^2}\text{ dx}$
$\Rightarrow\text{I}=\frac{1}{\sqrt{17}}\Big[\log\Big(\frac{\sqrt{17}+2\text{x}-1}{\sqrt{17}-2\text{x}+1}\Big)\Big]^2_0$
$\Rightarrow\text{I}=\frac{1}{\sqrt{17}}\Big\{\log\frac{\sqrt{17}+3}{\sqrt{17}-3}-\log\frac{\sqrt{17}-1}{\sqrt{17}+1}\Big\}$
$\Rightarrow\text{I}=\frac{1}{\sqrt{17}}\Big\{\log\frac{26+6\sqrt{17}}{8}-\log\frac{18-2\sqrt{17}}{16}\Big\}$
$\Rightarrow\text{I}=\frac{1}{\sqrt{17}}\Big\{\log\frac{52+12\sqrt{17}}{18-2\sqrt{17}}\Big\}$
$\Rightarrow\text{I}=\frac{1}{\sqrt{17}}\Big\{\log\frac{52+12\sqrt{17}}{18-2\sqrt{17}}\times\frac{18+2\sqrt{17}}{18+2\sqrt{17}}\Big\}$
$\Rightarrow\text{I}=\frac{1}{\sqrt{17}}\log\frac{1344+320\sqrt{17}}{256}$
$\Rightarrow\text{I}=\frac{1}{\sqrt{17}}\log\frac{21+5\sqrt{17}}{4}$

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