Question

If I = $\int_{0}^{\infty}\frac{\sqrt{x}dx}{(1 + x)(2 + x)(3 + x)}$, then I equals-

• A. $\frac{\pi}{2}$ (2$\sqrt{2}$– $\sqrt{3}$–1)
• B. $\frac{\pi}{2}$ (2$\sqrt{2}$+ $\sqrt{3}$–1)
• C. $\frac{\pi}{2}$ (2$\sqrt{2}$– $\sqrt{3}$+1)
• D. None of these

Answer 1

Answer: (A)

$\frac{\pi}{2}$ (2$\sqrt{2}$– $\sqrt{3}$–1)

Answer 2

Put $\sqrt{x}$= t or x = t², so that

I = 2$\int_{0}^{\infty}\frac{t^{2}}{(1 + t^{2})(2 + t^{2})(3 + t^{2})}$dt

= $\int_{0}^{\infty}\left( - \frac{1}{1 + t^{2}} + \frac{4}{2 + t^{2}} - \frac{3}{3 + t^{2}} \right)$dt

=$\left. \ \left( - \tan^{- 1}t + \frac{4}{\sqrt{2}}\tan^{- 1}\left( \frac{t}{\sqrt{2}} \right) - \frac{3}{\sqrt{3}}\tan^{- 1}\left( \frac{t}{\sqrt{3}} \right) \right) \right\rbrack_{0}^{\infty}$= –$\frac{\pi}{2}$ + 2$\sqrt { 2 }$ $\left( \frac{\pi}{2} \right)$– $\sqrt { 3 }$ $\left( \frac{\pi}{2} \right)$

= $\frac{\pi}{2}$ (2$\sqrt{2}$–$\sqrt{3}$–1)