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Indefinite Integrals — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsIndefinite Integrals4 Marks
Question
Evaluate the following intregals:
$\int\frac{\text{x}}{\sqrt{8+\text{x}-\text{x}^2}}\text{dx}$
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Answer

Let $\text{I}=\int\frac{\text{x}}{\sqrt{8+\text{x}-\text{x}^2}}\text{dx}$
Let $\text{x}=\lambda\frac{\text{d}}{\text{dx}}(8+\text{x}-\text{x}^2)+\mu$
$=\lambda(1-2\text{x})+{\mu}$
$\text{x}=(-2\lambda)\text{x}+\lambda+\mu$
compairing the coefficient of like powers of x,
$-2\lambda=1\ \Rightarrow\ \lambda=-\frac{1}{2}$
$\lambda+\mu=0\ \Rightarrow\Big(-\frac{1}{2}\Big)+\mu=0$
$\mu=\frac{1}{2}$
So, $\text{I}=\int\frac{-\frac{1}{2}(1-2\text{x})+\frac{1}{2}}{\sqrt{8+\text{x}-\text{x}^2}}\text{dx}$
$=-\frac{1}{2}\int\frac{(1-2\text{x})}{\sqrt{8+\text{x}-\text{x}^2}}\text{dx}+\frac{1}{2}\int\frac{1}{\sqrt{-\big[\text{x}^2-\text{x}-8\big]}}\text{dx}$
$=-\frac{1}{2}\int\frac{(1-2\text{x})}{\sqrt{8+\text{x}-\text{x}^2}}\text{dx}+\frac{1}{2}\int\frac{1}{\sqrt{-\Big[\text{x}^2-2\text{x}\big(\frac{1}{2}\big)+\big(\frac{1}{2}\big)^2-\big(\frac{1}{2}\big)^2-8}}\text{dx}$
$$$=-\frac{1}{2}\int\frac{(1-2\text{x})}{\sqrt{8+\text{x}-\text{x}^2}}\text{dx}+\frac{1}{2}\int\frac{1}{\sqrt{-\Big[\big(\text{x}-\frac{1}{2}\big)^2-\big(\frac{33}{4}\big)^2\Big]}}\text{dx}$
$\text{I}=-\frac{1}{2}\times2\sqrt{8+\text{x}-\text{x}^2}+\frac{1}{2}\sin^{-1}\Bigg(\frac{\text{x}-\frac{1}{2}}{\frac{\sqrt{33}}{2}}\Bigg)+\text{C}$ $\big[\text{since}, \int\frac{1}{\sqrt{\text{x}}}\text{dx}=2\sqrt{\text{x}}+\text{c},\int\frac{1}{\sqrt{\text{a}^2-\text{x}^2}}\text{dx}=\sin^{-1}\big(\frac{\text{x}}{\text{a}}\big)+\text{C}\big]$
$\text{I}=\sqrt{8+\text{x}-\text{x}^2}+\frac{1}{2}\sin^{-1}\Big(\frac{2\text{x}-1}{\sqrt{33}}\Big)+\text{C}$

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