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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

Similar Questions

To find the value of $\int \frac{1 + \log x}{x} dx$,the proper substitution is

$\int {x{e^{{x^2}}}} dx = $

$\int \frac{\cos^3 x}{\sin^2 x + \sin x} dx =$

What is the value of the integral $I = \int \frac{dx}{(1 + e^x)(1 + e^{-x})}$?

If $f(x) = \sqrt{\tan x}$ and $g(x) = \sin x \cdot \cos x$,then $\int \frac{f(x)}{g(x)} dx$ is equal to (where $C$ is a constant of integration).

Vedclass Test Series

Exam Paper Generator

Online Exam Module