The value of the integral [_- _e 2^ _e 2 e^x( _e(e^x+√1+e^2 x)) d x] is equal to :

Question

The value of the integral $$\int_\limits{-\log _{e} 2}^{\log _{e} 2} e^{x}\left(\log _{e}\left(e^{x}+\sqrt{1+e^{2 x}}\right)\right) d x$$ is equal to :

Accepted Answer

$$\int_\limits{-\log _{e} 2}^{\log _{e} 2} e^{x}\left(\log _{e}\left(e^{x}+\sqrt{1+e^{2 x}}\right)\right) d x$$

Let $e^x=t \Rightarrow e^x d x=d t$

When, $x \rightarrow-\log _e 2$, then $t \rightarrow \frac{1}{2}$

When, $x \rightarrow \log _e 2$, then $t \rightarrow 2$

$$
I=\int_\limits{\frac{1}{2}}^2\left[\log _e\left(t+\sqrt{1+t^2}\right)\right] d t
$$ ...........(i)

On applying integration by part method in Eq. (i), we get

$$
\begin{aligned}
& I=\left[t \log _e\left(t+\sqrt{1+t^2}\right)\right]_{1 / 2}^2-\int_{1 / 2}^2 \frac{t}{t+\sqrt{1+t^2}}\left(1+\frac{2 t}{2 \sqrt{1+t^2}}\right) d t \\
& =2 \log _e(2+\sqrt{5})-\frac{1}{2} \log _e\left(\frac{1+\sqrt{5}}{2}\right)-\int_{1 / 2}^2 \frac{t}{\sqrt{1+t^2}} d t
\end{aligned}
$$

$$
=\log _e\left(\frac{(2+\sqrt{5})^2}{\left(\frac{1+\sqrt{5}}{2}\right)^{1 / 2}}\right)-\frac{1}{2} \int_{1 / 2}^2 \frac{2 t}{\sqrt{1+t^2}} d t
$$ .............(ii)

Let $\quad I_1=\int_{1 / 2}^2 \frac{2 t}{\sqrt{1+t^2}} d t$

Let $1+t^2=w$

$2 t d t=d w$

When, $t \rightarrow \frac{1}{2}$, then $w=\frac{5}{4}$

When, $t \rightarrow 2$, then $w=5$

$$
\begin{aligned}
I_1 & =\int_{5 / 4}^5 \frac{1}{\sqrt{w}} d w \\
& =[2 \sqrt{w}]_{5 / 4}^5 \\
& =2\left[\sqrt{5}-\frac{\sqrt{5}}{2}\right]=\sqrt{5}
\end{aligned}
$$

On substitute value of $I_1$ in Eq. (ii), we get

$$
I=\log _e\left(\frac{\sqrt{2}(2+\sqrt{5})^2}{\sqrt{1+\sqrt{5}}}\right)-\frac{\sqrt{5}}{2}
$$

Solutions · 1