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Indefinite Integrals — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsIndefinite Integrals5 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 power of x,
$-2\lambda=3\ \Rightarrow\lambda=-\frac{3}{2}$
$-2\lambda+\mu=1\ \Rightarrow-2\Big(-\frac{3}{2}\Big)+\mu=1$
$\mu=-2$
So, $\text{I}=\int\frac{-\frac{3}{2}(-2-2\text{x})-2}{\sqrt{5-2\text{x}-\text{x}^2}}\text{ dx}$
$=-\frac{3}{2}\int\frac{(-2-2\text{x})}{\sqrt{5-2\text{x}-\text{x}^2}}\text{ dx}-2\int\frac{1}{\sqrt{-\big[\text{x}^2+2\text{x}-5\big]}}\text{ dx}$
$=-\frac{3}{2}\int\frac{(-2-2\text{x})}{\sqrt{5-2\text{x}-\text{x}^2}}\text{ dx}-2\int\frac{1}{\sqrt{-\big[\text{x}^2+2\text{x}+(1)^2-(1)^2-5\big]}}\text{ dx}$
$=-\frac{3}{2}\int\frac{(-2-2\text{x})}{\sqrt{5-2\text{x}-\text{x}^2}}\text{ dx}-2\int\frac{1}{\sqrt{-\big[(\text{x}+1)^2-(\sqrt{6})^2\big]}}\text{ dx}$
$=-\frac{3}{2}\int\frac{(-2-2\text{x})}{\sqrt{5-2\text{x}-\text{x}^2}}\text{ dx}-2\int\frac{1}{\sqrt{\big[(\sqrt{6})^2-(\text{x}+1)^2\big]}}\text{ dx}$
$\text{I}=-\frac{3}{2}\times2\sqrt{5-2\text{x}-\text{x}^2}-2\sin^{-1}\Big(\frac{\text{x}+1}{\sqrt{6}}\Big)+\text{c}$ $\big[\text{since}, \int\frac{1}{\sqrt{\text{x}}}\text{dx}=2\sqrt{\text{x}}+\text{c},\int\frac{1}{\sqrt{\text{a}^2-\text{x}^2}}\text{dx}=\sin^{-1}\big(\frac{\text{x}}{\text{a}}\big)+\text{c}\big]$
$\text{I}=-\frac{3}{2}\times2\sqrt{5-2\text{x}-\text{x}^2}-2\sin^{-1}\Big(\frac{\text{x}+1}{\sqrt{6}}\Big)+\text{C}$

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