An integral problem for Nihar

$\large \displaystyle \int_0^{\ln 4} \frac{e^t}{\sqrt{e^{2t} + 9}} .dt\\ \large \displaystyle \text{Let } y = e^t \implies dy = e^t . dt\\ \large \displaystyle \text{If } t = 0, y = 1; \text{If } t = \ln 4, y = 4.$

$\large \displaystyle \int_1^4 \frac{dy}{\sqrt{t^2 + 9}} = \int_1^4 \frac{dy}{\sqrt{t^2 + 3^2}}\\ \large \displaystyle = \left[\ln (y + \sqrt{y^2 + 3^2}) \right]_1^4\\ \large \displaystyle = \ln (4 + \sqrt{16+9}) - \ln (1 + \sqrt{1+9})\\ \large \displaystyle = \ln 9 - \ln (1 + \sqrt{10}) = \ln \frac{9}{1+\sqrt{10}} \approx \boxed{0.771}$

Since $\dfrac{d}{dx}\sinh^{-1} x = \dfrac{1}{\sqrt{x^2+1}}$ , note that by differentiation - Chain Rule , $\dfrac{d}{dx} \sinh^{-1} \dfrac{e^x}{3} = \dfrac{e^x}{\sqrt{e^{2x}+9}}$ which is exactly our integral! Thus we have $\displaystyle\int \dfrac{e^x}{\sqrt{e^{2x}+9}}=\sinh^{-1} \dfrac{e^x}{3}$ and substituting the boundaries we have $I=\sinh^{-1} \dfrac{4}{3} - \sinh^{-1} \dfrac{1}{3} \approx \boxed{0.771}$ .

Another method would be to use two substitutions: $y=e^t$ first and then $y=3\tan t$ .

$\begin{aligned} \int_{0}^{\ln(4)}\frac{e^t}{\sqrt{e^{2t}+9}}\,dx &= \sinh^{-1}\left( \frac{e^t}{3}\right) \quad\quad\quad\quad{\text{Indefinite integral}}\\&= \ln(3) - \ce{csch}^{-1}(3) \\&= 0.77116. . . \space \space \square\end{aligned}$

$\LARGE \text{ADIOS!!!}$