use an appropriate substitution and then a trigonometric substitution.\n int_{0}^{ln 4} \frac{e^{t}dt}{sqrt{e…

Explanation:

Step1: First - order substitution

Let $u = e^{t}$, then $du=e^{t}dt$. When $t = 0$, $u = e^{0}=1$; when $t=\ln4$, $u = e^{\ln4}=4$. The integral becomes $\int_{1}^{4}\frac{du}{\sqrt{u^{2}+25}}$.

Step2: Trigonometric substitution

Let $u = 5\tan\theta$, then $du=5\sec^{2}\theta d\theta$ and $\sqrt{u^{2}+25}=\sqrt{25\tan^{2}\theta + 25}=5\sec\theta$. When $u = 1$, $\tan\theta=\frac{1}{5}$, $\theta=\arctan\frac{1}{5}$; when $u = 4$, $\tan\theta=\frac{4}{5}$, $\theta=\arctan\frac{4}{5}$. The integral $\int_{1}^{4}\frac{du}{\sqrt{u^{2}+25}}$ becomes $\int_{\arctan\frac{1}{5}}^{\arctan\frac{4}{5}}\frac{5\sec^{2}\theta d\theta}{5\sec\theta}=\int_{\arctan\frac{1}{5}}^{\arctan\frac{4}{5}}\sec\theta d\theta$.

Step3: Integrate $\sec\theta$

The antiderivative of $\sec\theta$ is $\ln|\sec\theta+\tan\theta|$. So $\int_{\arctan\frac{1}{5}}^{\arctan\frac{4}{5}}\sec\theta d\theta=\left[\ln|\sec\theta+\tan\theta|\right]_{\arctan\frac{1}{5}}^{\arctan\frac{4}{5}}$. If $\tan\theta=\frac{4}{5}$, then $\sec\theta=\frac{\sqrt{4^{2}+5^{2}}}{5}=\frac{\sqrt{41}}{5}$; if $\tan\theta=\frac{1}{5}$, then $\sec\theta=\frac{\sqrt{1 + 25}}{5}=\frac{\sqrt{26}}{5}$. The result is $\ln\left(\frac{\sqrt{41}}{5}+\frac{4}{5}\right)-\ln\left(\frac{\sqrt{26}}{5}+\frac{1}{5}\right)=\ln\frac{\sqrt{41}+4}{\sqrt{26}+1}$.

Answer:

$\ln\frac{\sqrt{41}+4}{\sqrt{26}+1}$