int sqrt{e^x-1} , dx = | vedclass

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

Question

(C) Let $I = \int \sqrt{e^x-1} \, dx$. Substitute $e^x - 1 = t^2$,which implies $e^x = t^2 + 1$. Differentiating both sides,$e^x \, dx = 2t \, dt$. Th

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) Let $I = \int \sqrt{e^x-1} \, dx$. Substitute $e^x - 1 = t^2$,which implies $e^x = t^2 + 1$. Differentiating both sides,$e^x \, dx = 2t \, dt$. Thus,$dx = \frac{2t}{t^2+1} \, dt$. Substituting these into the integral: $I = \int t \cdot \frac{2t}{t^2+1} \, dt = \int \frac{2t^2}{t^2+1} \, dt$. Rewrite the numerator: $I = \int \frac{2(t^2+1) - 2}{t^2+1} \, dt = \int \left( 2 - \frac{2}{t^2+1} ight) \, dt$. Integrating term by term: $I = 2t - 2 	an^{-1}(t) + c$. Substituting $t = \sqrt{e^x-1}$ back: $I = 2 \sqrt{e^x-1} - 2 	an^{-1} \sqrt{e^x-1} + c$.

$\int \sqrt{e^x-1} \, dx =$

Similar Questions

The value of $\int \frac{d x}{x^2(x^4+1)^{\frac{3}{4}}}$ is

If $\int \frac{dx}{x (\sqrt{x^4 - 1})} = \frac{1}{k} \sec^{-1} (x^k)$,then the value of $k = $

$\int \sin^{\frac{-1}{2}}x \cos^{\frac{-7}{2}}x \, dx = $

$\int \frac{\sin^3 2x}{\cos^5 2x} \, dx = $

$\int \tan x \sec^2 x \sqrt{1 - \tan^2 x} \; dx = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module