int_{frac{2}{e}}^{frac{1}{e}} frac{1}{x(log x | vedclass

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

Question

(D) Let $I = \int_{\frac{2}{e}}^{\frac{1}{e}} \frac{1}{x(\log x)^{1/3}} dx$.Substitute $t = \log x$,then $dt = \frac{1}{x} dx$.When $x = \frac{1}{e}$,

Vedclass · Accepted Answer

$\frac{3}{2}\left\{1-(\log 2-1)^{\frac{2}{3}}\right\}$

Vedclass · Answer

(D) Let $I = \int_{\frac{2}{e}}^{\frac{1}{e}} \frac{1}{x(\log x)^{1/3}} dx$.Substitute $t = \log x$,then $dt = \frac{1}{x} dx$.When $x = \frac{1}{e}$,$t = \log(\frac{1}{e}) = -1$.When $x = \frac{2}{e}$,$t = \log(\frac{2}{e}) = \log 2 - \log e = \log 2 - 1$.Thus,$I = \int_{\log 2 - 1}^{-1} t^{-1/3} dt$.Integrating,we get $I = \left[ \frac{t^{2/3}}{2/3} \right]_{\log 2 - 1}^{-1} = \frac{3}{2} \left[ t^{2/3} \right]_{\log 2 - 1}^{-1}$.$I = \frac{3}{2} \left[ (-1)^{2/3} - (\log 2 - 1)^{2/3} \right]$.Since $(-1)^{2/3} = ((-1)^2)^{1/3} = 1^{1/3} = 1$,we have $I = \frac{3}{2} \left[ 1 - (\log 2 - 1)^{2/3} \right]$.

$\int_{\frac{2}{e}}^{\frac{1}{e}} \frac{1}{x(\log x)^{\frac{1}{3}}} dx$ is equal to

Similar Questions

$\int_0^3 \frac{dx}{(x+2) \sqrt{x+1}} = $

Evaluate the integral $\int_{0}^{2} x \sqrt{x+2} \, dx$ using the substitution $x+2=t^{2}$.

It is given that $\frac{d}{dt}(t \log t - t) = \log t$. Then, $\exp \left( \int_0^1 2x \log(1+x^2) dx \right) = $

$\int_0^1 \frac{1}{2+\sqrt{x}} \, dx =$

Evaluate the following integral:
$\int_{4}^{9} \frac{\sqrt{x}}{\left(30-x^{\frac{3}{2}}\right)^{2}} d x$

Vedclass Test Series

Exam Paper Generator

Online Exam Module