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

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

Question

(A) Let $t = x + \log \sec x$. Then,the derivative of $t$ with respect to $x$ is: $\frac{dt}{dx} = \frac{d}{dx}(x) + \frac{d}{dx}(\log \sec x) = 1 + \

Vedclass · Accepted Answer

$\log (x + \log \sec x) + c$

Vedclass · Answer

(A) Let $t = x + \log \sec x$. Then,the derivative of $t$ with respect to $x$ is: $\frac{dt}{dx} = \frac{d}{dx}(x) + \frac{d}{dx}(\log \sec x) = 1 + \frac{1}{\sec x} \cdot (\sec x 	an x) = 1 + 	an x$. So,$dt = (1 + 	an x) \, dx$. Substituting these into the integral: $\int \frac{1 + 	an x}{x + \log \sec x} \, dx = \int \frac{1}{t} \, dt = \log |t| + c$. Replacing $t$ with the original expression: $\log |x + \log \sec x| + c$.

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

Similar Questions

$\int \frac{10^{\frac{x}{2}}}{\sqrt{10^{-x}-10^x}} dx=$

$\int \frac{1}{x+x \log x} d x=$ . . . . . . .

$\int \frac{\sin 2x}{4 \sin^2 x + 9 \cos^2 x} \, dx = $ (Where $C$ is a constant of integration).

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

$\int \frac{1+2 e^{-x}}{1-2 e^{-x}} d x=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module