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Indefinite Integrals — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsIndefinite Integrals4 Marks
Question
Evaluate the following intregals:
$\int\frac{3\text{x}+1}{\sqrt{5-2\text{x}-\text{x}^2}}\text{dx}$
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Answer

 Let  $\text{I}=\int\frac{3\text{x}+1}{\sqrt{5-2\text{x}-\text{x}^2}}\text{dx}$
let $3\text{x}+1=\lambda\frac{\text{d}}{\text{dx}}(5-2\text{x}-\text{x}^2)+\mu$
$=\lambda(-2-2\text{x})+\mu$
$3\text{x}+1=(-2\lambda)\text{x}-2\lambda+\mu$
Compairing the coefficient of like powewrs of x,
$-2\lambda=3\ \Rightarrow\lambda=-\frac{3}{2}$
$2\lambda+\mu=1\ \Rightarrow -2\big(-\frac{3}{2}\big)+\mu=1$
$\Rightarrow\mu=-2$
So, $\text{I}=\int\frac{-\frac{3}{2}(-2-2\text{x})-2}{\sqrt{5-2\text{x}-\text{x}^2}}\text{dx}$
$-\frac{3}{2}\int\frac{(-2-2\text{x})}{\sqrt{5-2\text{x}-\text{x}^2}}\text{dx}-3\int\frac{1}{\sqrt{-\Big[\text{x}^2+2\text{x}-5\Big]}}\text{dx}$
$=-\frac{3}{2}\int\frac{(-2-2\text{x})}{\sqrt{5-2\text{x}-\text{x}}^2}\text{dx}-2\int\frac{1}{\sqrt{-\big[\text{x}^2+2\text{x}+(1)^2-(1)^2+5\big]}}\text{dx}$
$=-\frac{3}{2}\int\frac{(-2-2\text{x})}{\sqrt{5-2\text{x}-\text{x}}^2}\text{dx}-2\int\frac{1}{\sqrt{-\big[(\text{x}+1)^2-(\sqrt{6})^2\big]}}\text{dx}$
$=-\frac{3}{2}\int\frac{(-2-2\text{x})}{\sqrt{5-2\text{x}-\text{x}}^2}\text{dx}-2\int\frac{1}{\sqrt{(\sqrt{6})^2-(\text{x}+1)^2}}\text{dx}$
$\text{I}=-\frac{3}{2}\times2\sqrt{5-2\text{x}-\text{x}^{-2}}\ 2\sin^{-1}\Big(\frac{\text{x}+1}{\sqrt{6}}\Big)+\text{C}$ $\big[\text{since},\int\frac{1}{\sqrt{\text{x}}}\text{dx}=2\sqrt{\text{x}}+\text{C},\int\frac{1}{\sqrt{\text{a}^2-\text{x}^2}}\text{dx}=\sin^{-1}\big(\frac{\text{x}}{\text{a}}\big)+\text{C}\big]$
$\text{I}=-3\sqrt{5-2\text{x}-\text{x}^2}-2\sin^{-1}\Big(\frac{\text{x}+1}{\sqrt{6}}\Big)+\text{C}$

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