int frac{dx}{xsqrt{1 - (log x)^2}} = | vedclass

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

Question

(C) Let $I = \int \frac{dx}{x\sqrt{1 - (\log x)^2}}$.Substitute $t = \log x$.Then,the derivative is $dt = \frac{1}{x} dx$.Substituting these into the

Vedclass · Accepted Answer

$\sin^{-1}(\log x) + c$

Vedclass · Answer

(C) Let $I = \int \frac{dx}{x\sqrt{1 - (\log x)^2}}$.Substitute $t = \log x$.Then,the derivative is $dt = \frac{1}{x} dx$.Substituting these into the integral,we get:$I = \int \frac{dt}{\sqrt{1 - t^2}}$.Using the standard integral formula $\int \frac{dt}{\sqrt{1 - t^2}} = \sin^{-1}(t) + c$,we obtain:$I = \sin^{-1}(t) + c$.Substituting back $t = \log x$,we get:$I = \sin^{-1}(\log x) + c$.

$\int \frac{dx}{x\sqrt{1 - (\log x)^2}} = $

Similar Questions

If $\int \frac{x^{\frac{1}{2}}}{\sqrt{a^3-x^3}} dx = P(x) + c$,then $P(x) =$

The value of $\int \frac{x^{2} d x}{\sqrt{x^{6}+a^{6}}}$ is equal to

$\int \frac{dx}{x[(\log x)^2 + 4\log x - 1]} = $

Find the following integral: $\int \frac{\sin x}{\sin (x+a)} d x$

If $I = \int \frac{2x-7}{\sqrt{3x-2}} \, dx$,then $I$ is given by

Vedclass Test Series

Exam Paper Generator

Online Exam Module