The value of the integral _-_e2^_e2e^x_e(e^x+√1+e^2x)dx is equal to:

Question

The value of the integral $\displaystyle\int_{-\log_e2}^{\log_e2}e^x\log_e\left(e^x+\sqrt{1+e^{2x}}\right)dx$ is equal to:

Accepted Answer

Let $e^x=t$, $e^x\,dx=dt$; limits become $t=1/2$ to $t=2$:

$$I = \int_{1/2}^2\ln\left(t+\sqrt{1+t^2}\right)dt$$

Integrating by parts:

$$I = \Big[t\ln(t+\sqrt{1+t^2})\Big]_{1/2}^2 - \int_{1/2}^2\frac{t}{\sqrt{1+t^2}}\,dt$$

$$= 2\ln(2+\sqrt5)-\frac12\ln\left(\frac{1+\sqrt5}2\right) - \Big[\sqrt{1+t^2}\Big]_{1/2}^2$$

$$= 2\ln(2+\sqrt5)-\frac12\ln\left(\frac{1+\sqrt5}2\right) - \left(\sqrt5-\frac{\sqrt5}2\right)$$

$$= \log_e\left(\frac{\sqrt2(2+\sqrt5)^2}{1+\sqrt5}\right)-\frac{\sqrt5}2$$

Solutions · 1