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Question

Question: \[\int_{0}^{\infty}\frac{dx}{\left( x + \sqrt{x^{2} + 1} \right)^{3}} =\]...

∫0∞dx(x+x2+1)3=\int_{0}^{\infty}\frac{dx}{\left( x + \sqrt{x^{2} + 1} \right)^{3}} =

A

38\frac{3}{8}

B

18\frac{1}{8}

C

−38- \frac{3}{8}

D

None of these

Answer

38\frac{3}{8}

Explanation

Solution

Puttingx=tan⁡θx = \tan\theta, we get ∫0∞dx(x+x2+1)3\int_{0}^{\infty}\frac{dx}{\left( x + \sqrt{x^{2} + 1} \right)^{3}}

=∫0π/2sec⁡2θdθ(tan⁡θ+sec⁡θ)3=∫0π/2cos⁡θ(1+sin⁡θ)3dθ= \int_{0}^{\pi/2}\frac{\sec^{2}\theta d\theta}{(\tan\theta + \sec\theta)^{3}} = \int_{0}^{\pi/2}{\frac{\cos\theta}{(1 + \sin\theta)^{3}}d\theta}

=[−12(1+sin⁡θ)2]0π/2=−18+12=38= \left\lbrack - \frac{1}{2(1 + \sin\theta)^{2}} \right\rbrack_{0}^{\pi/2} = - \frac{1}{8} + \frac{1}{2} = \frac{3}{8}.