If int sqrt{x-frac{1}{x}}left(frac{x^{2}+1}{x | vedclass

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(B) Let $I = \int \sqrt{x-\frac{1}{x}}\left(\frac{x^{2}+1}{x^{2}}\right) d x$. Substitute $u = x - \frac{1}{x}$. Then $du = (1 + \frac{1}{x^2}) dx = \

Vedclass · Accepted Answer

$\frac{3}{2}$

Vedclass · Answer

(B) Let $I = \int \sqrt{x-\frac{1}{x}}\left(\frac{x^{2}+1}{x^{2}}\right) d x$. Substitute $u = x - \frac{1}{x}$. Then $du = (1 + \frac{1}{x^2}) dx = \frac{x^2+1}{x^2} dx$. Substituting these into the integral,we get: $I = \int \sqrt{u} du = \int u^{1/2} du$. Integrating,we get: $I = \frac{u^{3/2}}{3/2} + c = \frac{2}{3} u^{3/2} + c$. Substituting back $u = x - \frac{1}{x}$: $I = \frac{2}{3} (x - \frac{1}{x})^{3/2} + c$. Comparing this with the given expression $\frac{2}{3}(x-\frac{1}{x})^k + c$,we find $k = \frac{3}{2}$.

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

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