int frac{1}{xsqrt{1 + log x}} , dx = | vedclass

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

Question

(C) Let $t = 1 + \log x$. Then,the derivative is $dt = \frac{1}{x} \, dx$. Substituting these into the integral: $\int \frac{1}{\sqrt{t}} \, dt = \int

Vedclass · Accepted Answer

$2\sqrt{1 + \log x} + c$

Vedclass · Answer

(C) Let $t = 1 + \log x$. Then,the derivative is $dt = \frac{1}{x} \, dx$. Substituting these into the integral: $\int \frac{1}{\sqrt{t}} \, dt = \int t^{-1/2} \, dt$. Using the power rule for integration,$\int t^n \, dt = \frac{t^{n+1}}{n+1} + c$: $= \frac{t^{1/2}}{1/2} + c = 2t^{1/2} + c$. Substituting $t = 1 + \log x$ back,we get: $= 2\sqrt{1 + \log x} + c$.

$\int \frac{1}{x\sqrt{1 + \log x}} \, dx = $

Similar Questions

Integrate the function: $\frac{\sec ^{2} x}{\sqrt{\tan ^{2} x+4}}$

$\int (x + 3)({x^2} + 6x + 10)^9 \, dx$ equals

Integrate the function $\frac{(x+1)(x+\log x)^{2}}{x}$.

$\int \frac{(x + 1)(x + \log x)^2}{x} \, dx = $

$\int \frac{1}{\sqrt{x}} \sin \sqrt{x} \, dx = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module