Question

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

• A. $log\left|1+\sqrt{2+x^{2}}\right|+c$
• B. $-log\left|1+\sqrt{2-x^{2}}\right|+c$
• C. $-x\,log\left|1-\sqrt{2-x^{2}}\right|+c$
• D. $x\,log\left|1-\sqrt{2+x^{2}}\right|+c$

Answer 1

The Correct Option is B

To solve the integral $\int \frac{xdx}{2-x^{2}+\sqrt{2-x^{2}}}$, we can employ a suitable substitution to simplify the expression inside the integral.

1. Let us consider the substitution $u = \sqrt{2-x^{2}}$. Then, differentiating both sides with respect to $x$, we get: $$du = \frac{-x}{\sqrt{2-x^2}} \, dx \Rightarrow x \, dx = -u \, du$$.
2. By substituting $u = \sqrt{2-x^{2}}$ and $x \, dx = -u \, du$ into the integral, the expression becomes: $$\int \frac{-u \, du}{u^2 + u}$$.
3. Simplify the integral: $$\int \frac{-u \, du}{u^2 + u} = \int \frac{-1 \, du}{u + 1}$$
   • This is a straightforward logarithmic integration, giving us: $$-\ln|u + 1| + C$$.
4. Substitute back $u = \sqrt{2-x^{2}}$: $$-\ln\left|1+\sqrt{2-x^{2}}\right| + C$$.

The final answer, after evaluating the integral, is: $-\ln\left|1+\sqrt{2-x^{2}}\right| + c$.

Among the given options, the correct answer is: $-\ln\left|1+\sqrt{2-x^{2}}\right|+c$, which matches the solution we derived.