Question

The value of ${\int }_1^{\infty }\frac{dx}{x^2\sqrt{1+x}}$ is

• A. $\sqrt{2}-\log (\sqrt{2}+1)$
• B. $\sqrt{2}\log (\sqrt{2}+1)$
• C. $\sqrt{2}+\log (\sqrt{2}+1)$
• D. none of these

Answer 1

Answer: (A)

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

Let $I={\int }_1^{\infty }\frac{dx}{x^2\sqrt{1+x}}$
Put $1+x=t^2\Rightarrow dx=2tdt$
$I={\int }_{\sqrt{2}}^{\infty }\frac{2t}{{(t^2-1)}^{2}t}dt$
$={[-\frac{1}{(t^2-1)}]}_{\sqrt{2}}^{\infty }-{\int }_{\sqrt{2}}^{\infty }\frac{dt}{(t^2-1)t^2}$
$=\frac{1}{\sqrt{2}}-{\int }_{\sqrt{2}}^{\infty }(\frac{1}{t^2-1}-\frac{1}{t^2})dt$
$=\frac{1}{\sqrt{2}}-{[\frac{1}{2}\log \left(\frac{t-1}{t+1}\right)+\frac{1}{t}]}_{\sqrt{2}}^{\infty }=\sqrt{2}-\log (\sqrt{2}-1)$