$\int \sqrt{x-\frac{1}{x}}\left(\frac{x^{2}+1}{x^{2}}\right) d x=\frac{2}{3}\left(x-\frac{1}{x}\right)^{k}+c \quad$, then value of k is

Question

Aicharya · Accepted Answer

$\begin{array}{l} 	ext { Let } \mathrm{I}=\int \sqrt{\mathrm{x}-\frac{1}{\mathrm{x}}}\left(\frac{\mathrm{x}^{2}+1}{\mathrm{x}^{2}}ight) \mathrm{d} \mathrm{x}=\frac{2}{3}\left(\mathrm{x}-\frac{1}{\mathrm{x}}ight)^{\mathrm{k}}+\mathrm{c} \ 	ext { Put } \sqrt{\mathrm{x}-\frac{1}{\mathrm{x}}}=\mathrm{t} \Rightarrow \frac{1}{2 \sqrt{\mathrm{x}-\frac{1}{\mathrm{x}}}}\left(1+\frac{1}{\mathrm{x}^{2}}ight) \mathrm{dx}=\mathrm{dt} \ 	herefore\left(\frac{\mathrm{x}^{2}+1}{\mathrm{x}^{2}}ight) \mathrm{d} \mathrm{x}=2 \mathrm{t} \mathrm{dt} \ 	herefore \mathrm{I}=\int \mathrm{t}(2 \mathrm{t}) \mathrm{dt}=2 \int \mathrm{t}^{2} \mathrm{dt}=\frac{2 \mathrm{t}^{3}}{3}=\frac{2}{3}\left[\sqrt{\mathrm{x}-\frac{1}{\mathrm{x}}}ight]^{3} \ =\frac{2}{3}\left[\left(\mathrm{x}-\frac{1}{\mathrm{x}}ight)^{\frac{1}{2}}ight]^{3}=\frac{2}{3}\left[\mathrm{x}-\frac{1}{\mathrm{x}}ight]^{\frac{3}{2}} \Rightarrow \mathrm{k}=\frac{3}{2} \end{array}$

$\int \sqrt{x-\frac{1}{x}}\left(\frac{x^{2}+1}{x^{2}}\right) d x=\frac{2}{3}\left(x-\frac{1}{x}\right)^{k}+c…

Solution