The integral $\int \frac{x d x}{2-x^2+\sqrt{2-x^2}}$ equals :
The integral $\int \frac{x d x}{2-x^2+\sqrt{2-x^2}}$ equals :
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$\log \left|1+\sqrt{2+x^2}\right|+c$
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$-\log \left|1+\sqrt{2-x^2}\right|+c$
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$-x \log \left|1-\sqrt{2-x^2}\right|+c$
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$x \log \left|1-\sqrt{2+x^2}\right|+c$
Solution
$\quad \mathrm{I}=\int \frac{x d x}{2-x^2+\sqrt{2-x^2}}$
$
\begin{aligned}
& \text { Put } t=\sqrt{2-x^2}, \frac{d t}{d x}=\frac{1}{2 \sqrt{2-x^2}} \cdot(-2 x) \\
& \begin{aligned}
& \Rightarrow-t d t=x d x \\
& \therefore \\
& \mathrm{I}=\int \frac{(-t) d t}{t^2+t}=-\int \frac{1}{t+1} d t=-\log |t+1| \\
&=-\log \left|\sqrt{2-x^2}+1\right|+c
\end{aligned}
\end{aligned}
$
Asked in: JEE Main 2013 (23 Apr Online)
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