# Chapter 9 - Further Applications of the Integral and Taylor Polynomials - 9.1 Arc Length and Surface Area - Exercises - Page 469: 27

$$\sqrt{1+e^{2 a}}+\frac{1}{2} \ln \frac{\sqrt{1+e^{2 a}}-1}{\sqrt{1+e^{2 a}}+1}-\sqrt{2}+\ln (1+\sqrt{2})$$

#### Work Step by Step

Since $$\int_{0}^{a} \sqrt{1+e^{2 x}} d x$$ Let $$u= \sqrt{1+e^{2 x}} \to du= \frac{e^{2 x}}{ \sqrt{1+e^{2 x}}}dx\to dx=\frac{u^2}{u^2-1}du$$ Then \begin{aligned} \int_{0}^{a} \sqrt{1+e^{2 x}} d x &=\int_{x=0}^{x=a} u \cdot \frac{u}{u^{2}-1} d u=\int_{x=0}^{x=a} \frac{u^{2}}{u^{2}-1} d u=\int_{x=0}^{x=a}\left(1+\frac{1}{u^{2}-1}\right) d u \\ &=\int_{x=0}^{x=a}\left(1+\frac{1}{2} \frac{1}{u-1}-\frac{1}{2} \frac{1}{u+1}\right) d u=\left.\left(u+\frac{1}{2} \ln (u-1)-\frac{1}{2} \ln (u+1)\right)\right|_{x=0} ^{x=a} \\ &=\left.\left[\sqrt{1+e^{2 x}}+\frac{1}{2} \ln \left(\frac{\sqrt{1+e^{2 x}}-1}{\sqrt{1+e^{2 x}}+1}\right)\right]\right|_{0} ^{a} \\ &=\sqrt{1+e^{2 a}}+\frac{1}{2} \ln \frac{\sqrt{1+e^{2 a}}-1}{\sqrt{1+e^{2 a}}+1}-\sqrt{2}+\frac{1}{2} \ln \frac{1+\sqrt{2}}{\sqrt{2}-1} \\ &=\sqrt{1+e^{2 a}}+\frac{1}{2} \ln \frac{\sqrt{1+e^{2 a}}-1}{\sqrt{1+e^{2 a}}+1}-\sqrt{2}+\ln (1+\sqrt{2}) \end{aligned}

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