int frac{d x}{sqrt{2 e^x-1}}= | vedclass

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(A) Let $I = \int \frac{d x}{\sqrt{2 e^x-1}}$.Substitute $u = \sqrt{2 e^x-1}$. Then $u^2 = 2 e^x - 1$,which implies $2 e^x = u^2 + 1$,or $e^x = \frac{

Vedclass · Accepted Answer

$2 \operatorname{Tan}^{-1}\left(\sqrt{2 e^x-1}\right)+c$

Vedclass · Answer

(A) Let $I = \int \frac{d x}{\sqrt{2 e^x-1}}$.Substitute $u = \sqrt{2 e^x-1}$. Then $u^2 = 2 e^x - 1$,which implies $2 e^x = u^2 + 1$,or $e^x = \frac{u^2+1}{2}$.Taking the natural logarithm on both sides,$x = \ln\left(\frac{u^2+1}{2}\right)$.Differentiating with respect to $u$,$dx = \frac{1}{\frac{u^2+1}{2}} \cdot \frac{2u}{2} du = \frac{2u}{u^2+1} du$.Substituting these into the integral:$I = \int \frac{1}{u} \cdot \frac{2u}{u^2+1} du = \int \frac{2}{u^2+1} du$.Integrating,we get $I = 2 \operatorname{Tan}^{-1}(u) + c$.Substituting back $u = \sqrt{2 e^x-1}$,we get $I = 2 \operatorname{Tan}^{-1}\left(\sqrt{2 e^x-1}\right) + c$.

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

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