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Question: If I = \(\int_{1}^{\infty}\frac{dx}{x^{2}\sqrt{1 + x}}\), then I equals...

If I = ∫1∞dxx21+x\int_{1}^{\infty}\frac{dx}{x^{2}\sqrt{1 + x}}, then I equals

A

2\sqrt{2}+ log (2\sqrt{2}–1)

B

2\sqrt{2}–log (2\sqrt{2}–1)

C

log (2\sqrt{2} + 1) + 2\sqrt{2}

D

2\sqrt{2} + log (3 –2\sqrt{2})

Answer

log (2\sqrt{2} + 1) + 2\sqrt{2}

Explanation

Solution

Put 1 + x = t2, then

I = 1t\frac{1}{t}dt

Integrating by parts, we get

I = – 1(t2−1)t]2∞\left. \ \frac{1}{(t^{2} - 1)t} \right\rbrack_{\sqrt{2}}^{\infty}– ∫2∞dt(t2−1)t2\int_{\sqrt{2}}^{\infty}\frac{dt}{(t^{2} - 1)t^{2}}

= 12\frac{1}{\sqrt{2}}– ∫2∞[1t2−1−1t2]\int_{\sqrt{2}}^{\infty}\left\lbrack \frac{1}{t^{2} - 1} - \frac{1}{t^{2}} \right\rbrackdt

= 12\frac{1}{\sqrt{2}}–  (12log⁡∣t−1t+1∣+1t)]2∞\left. \ \left( \frac{1}{2}\log\left| \frac{t - 1}{t + 1} \right| + \frac{1}{t} \right) \right\rbrack_{\sqrt{2}}^{\infty}

= 12\frac{1}{\sqrt{2}}– 12\frac{1}{2}log  ∣1−1/t1+1/t∣]t=∞\left. \ \left| \frac{1 - 1/t}{1 + 1/t} \right| \right\rbrack_{t = \infty}+ 0

+ 12\frac{1}{2}log ∣2−12+1∣\left| \frac{\sqrt{2} - 1}{\sqrt{2} + 1} \right|+12\frac{1}{\sqrt{2}}

= 2\sqrt{2}– log (2\sqrt{2}–1)