The integral xdx 2 - x^2 + 2 - x^2 equals - Vidyadip

MCQ

The integral $\int {\frac{{xdx}}{{2 - {x^2} + \sqrt {2 - {x^2}} }}} $ equals

A
$ \log \left| {1 + \sqrt {2 + {x^2}} } \right| + c$
✓
$ - \log \left| {1 + \sqrt {2 - {x^2}} } \right| + c$
C
$ - x\log \left| {1 - \sqrt {2 - {x^2}} } \right| + c$
D
$ x\log \left| {1 - \sqrt {2 + {x^2}} } \right| + c$

✓

Answer

Correct option: B.

$ - \log \left| {1 + \sqrt {2 - {x^2}} } \right| + c$

b
$\mathrm{I}=\int \frac{x d x}{2-x^{2}+\sqrt{2-x^{2}}}$

Put $t=\sqrt{2-x^{2}}, \frac{d t}{d x}$

$=\frac{1}{2 \sqrt{2-x^{2}}} \cdot(-2 x)$

$\Rightarrow-t d t=x d x$

$\therefore $

$I=\int \frac{(-t) d t}{t^{2}+t}$

$=-\int \frac{1}{t+1} d t$

$=-\log |t+1|$

$=-\log |\sqrt{2-x^{2}}+1|+c$

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