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Mathematics Question on General and Particular Solutions of a Differential Equation

The integral ∫xdx2−x2+2−x2\int \frac{xdx}{2-x^{2}+\sqrt{2-x^{2}}} equals

A

log∣1+2+x2∣+clog\left|1+\sqrt{2+x^{2}}\right|+c

B

−log∣1+2−x2∣+c-log\left|1+\sqrt{2-x^{2}}\right|+c

C

−x log∣1−2−x2∣+c-x\,log\left|1-\sqrt{2-x^{2}}\right|+c

D

x log∣1−2+x2∣+cx\,log\left|1-\sqrt{2+x^{2}}\right|+c

Answer

−log∣1+2−x2∣+c-log\left|1+\sqrt{2-x^{2}}\right|+c

Explanation

Solution

I=∫x dx2−x2+2−x2I=\int\frac{ x\,dx}{2-x^{2}+\sqrt{2-x^{2}}}
Put t=2−x2,dtdx=122−x2.(−2x)t=\sqrt{2-x^{2} }, \frac{dt}{dx}=\frac{1}{2\sqrt{2-x^{2}}}.\left(-2x\right)
⇒−t dt=x dx\Rightarrow -t\,dt=x\,dx
∴I=∫(−t)dtt2+t=−∫1t+1dt=−log∣t+1∣\therefore I=\int \frac{\left(-t\right)dt}{t^{2}+t}=-\int \frac{1}{t+1}dt=-log\left|t+1\right|
=−log∣2−x2+1∣+c=-log\left|\sqrt{2-x^{2}}+1\right|+c