Topic: Engineering Mathematics
Evaluate $\int_{0}^{1}\int_{0}^{\sqrt{1+x^{2}}}\frac{dy\cdot dx}{(1+x^{2}+y^{2})}$
- A. $\frac{\pi}{2}[\log(1+\sqrt{2})]$
- B. $\frac{\pi}{4}[\log(1+\sqrt{2})]$
- C. $\frac{\pi}{2}[\log(1-\sqrt{2})]$
- D. $\frac{\pi}{4}[\log(1-\sqrt{2})]$
Answer: B. $\frac{\pi}{4}[\log(1+\sqrt{2})]$
Explanation: Let $I = \int_{0}^{1} \left[ \int_{0}^{\sqrt{1+x^{2}}} \frac{dy}{(1+x^{2})+y^{2}} \right] dx$.
Using $\int \frac{1}{a²+u²}du = \frac{1}{a}\tan^{-1}(\frac{u}{a})$, where $a = \sqrt{1+x²}$:
$I = \int_{0}^{1} \left[ \frac{1}{\sqrt{1+x²}} \tan^{-1}\left(\frac{y}{\sqrt{1+x²}}\right) \right]_0^{\sqrt{1+x²}} dx = \int_0^1 \frac{1}{\sqrt{1+x²}} [\tan^{-1}(1) - \tan^{-1}(0)] dx$.
$I = \frac{\pi}{4} \int_0^1 \frac{1}{\sqrt{1+x²}} dx = \frac{\pi}{4} [\ln(x+\sqrt{1+x²})]_0^1 = \frac{\pi}{4} \ln(1+\sqrt{2})$.