int {frac{{{e^{{{tan }^{ - 1}}sqrt x }}}}{{sq | vedclass

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

Question

(D) Let $t = 	an^{-1} \sqrt{x}$.Differentiating both sides with respect to $x$,we get:$\frac{dt}{dx} = \frac{1}{1 + (\sqrt{x})^2} 	imes \frac{1}{2\s

Vedclass · Accepted Answer

$2{e^{{{	an }^{ - 1}}\sqrt x }} + c$

Vedclass · Answer

(D) Let $t = 	an^{-1} \sqrt{x}$.Differentiating both sides with respect to $x$,we get:$\frac{dt}{dx} = \frac{1}{1 + (\sqrt{x})^2} 	imes \frac{1}{2\sqrt{x}} = \frac{1}{1+x} 	imes \frac{1}{2\sqrt{x}}$.Thus,$dt = \frac{dx}{2\sqrt{x}(1+x)} = \frac{dx}{2(\sqrt{x} + x\sqrt{x})}$.This implies $\frac{dx}{\sqrt{x} + x\sqrt{x}} = 2 dt$.Substituting these into the integral:$\int e^t (2 dt) = 2 \int e^t dt = 2e^t + c$.Replacing $t$ with $	an^{-1} \sqrt{x}$,we get:$2e^{	an^{-1} \sqrt{x}} + c$.

$\int {\frac{{{e^{{{\tan }^{ - 1}}\sqrt x }}}}{{\sqrt x + x\sqrt x }}} dx = $

Similar Questions

$\int x^2 (3)^{x^3 + 1} dx = $

$\int \frac{\sec^2 x}{\log ((\tan x)^{\tan x})} dx = $

Integrate the function $\left(x^{3}-1\right)^{\frac{1}{3}} x^{5}$.

Integrate the function $\frac{x^{3} \sin \left(\tan ^{-1} x^{4}\right)}{1+x^{8}}$.

If $\int \frac{d x}{\sqrt{\sin ^3 x \cos x}}=g(x)+c$,then $g(x)$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module