Question

${\int }_1^2{(x+\frac{1}{x})}^{3/2}\frac{x^2-1}{x^2}dx$

• A. $\frac{5}{2}\sqrt{\left(\frac{5}{2}\right)}+\frac{8}{5}\sqrt{2}$
• B. $\frac{5}{2}\sqrt{\left(\frac{5}{2}\right)}-\frac{8}{5}\sqrt{2}$
• C. $\sqrt{\left(\frac{5}{2}\right)}-\frac{8}{5}\sqrt{2}$
• D. $\frac{3}{2}\sqrt{\left(\frac{3}{2}\right)}-\frac{8}{5}\sqrt{2}$

Answer 1

The correct option is B $\frac{5}{2}\sqrt{\left(\frac{5}{2}\right)}-\frac{8}{5}\sqrt{2}$

Let $I={\int }_1^2{(x+\frac{1}{x})}^{3/2}\frac{x^2-1}{x^2}dx$
Put $x+\frac{1}{x}=t\Rightarrow (1-\frac{1}{x^2})dx\Rightarrow dt$
Therefore
$I={\int }_2^{5/2}{t}^{3/2}dt=\frac{2}{5}{t}^{5/2}=\frac{2}{5}[{\left(\frac{5}{2}\right)}^{5/2}-2^{5/2}]$
$=\frac{2}{5}[{\left(\frac{5}{2}\right)}^2\sqrt{\left(\frac{5}{2}\right)}-2^2\sqrt{2}]=\frac{5}{2}\sqrt{\left(\frac{5}{2}\right)}-\frac{8}{5}\sqrt{2}$
Hence, option 'B' is correct.