Question

$\int {\left(x+\frac{1}{x}\right)}^{3/2}\left(\frac{x^2-1}{x^2}\right)dx$ is equal to

• A. $\frac{5}{2}{\left(x+\frac{1}{x}\right)}^{5/2}+c$
• B. $\frac{2}{5}{\left(x+\frac{1}{x}\right)}^{5/2}+c$
• C. $2{\left(x+\frac{1}{x}\right)}^{1/2}+c$
• D. None of these

Answer 1

The correct option is B $\frac{2}{5}{\left(x+\frac{1}{x}\right)}^{5/2}+c$

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

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

Let $t=x+\frac{1}{x}$

$dt=(1-\frac{1}{x^2})dx$

$I=\int (t{)}^{3/2}dt$

$I=\frac{(t{)}^{5/2}}{5/2}+c$

$I=\frac{2}{5}(x+\frac{1}{x}{)}^{5/2}+c$