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

Question

The value of the integral $\int _{-{\log }_{e}2}^{{\log }_{e}2}{e}^{x}({\log }_{e}({e}^{x}+\sqrt{1+{e}^{2x}}))dx$ is equal to :

Accepted Answer

$\int _{-{\log }_{e}2}^{{\log }_{e}2}{e}^{x}({\log }_{e}({e}^{x}+\sqrt{1+{e}^{2x}}))dx$

Let ${e}^{x}=t\Rightarrow {e}^{x}dx=dt$

When, $x	o -{\log }_{e}2$, then $t	o \frac{1}{2}$

When, $x	o {\log }_{e}2$, then $t	o 2$

$I=\int _{\frac{1}{2}}^{2}[{\log }_{e}(t+\sqrt{1+{t}^{2}})]dt$ ...........(i)

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

$\begin{gathered} I={[t{\log }_{e}(t+\sqrt{1+{t}^{2}})]}_{2}^{1/2}-{\int }_{2}^{1/2}\frac{t}{t+\sqrt{1+{t}^{2}}}(1+\frac{2t}{2\sqrt{1+{t}^{2}}})dt \ =2{\log }_{e}(2+\sqrt{5})-\frac{1}{2}{\log }_{e}(\frac{1+\sqrt{5}}{2})-{\int }_{2}^{1/2}\frac{t}{\sqrt{1+{t}^{2}}}dt \end{gathered}$

$={\log }_{e}(\frac{(2+\sqrt{5}{)}^{2}}{{(\frac{1+\sqrt{5}}{2})}^{1/2}})-\frac{1}{2}{\int }_{2}^{1/2}\frac{2t}{\sqrt{1+{t}^{2}}}dt$ .............(ii)

Let $\, {I}_{1}={\int }_{2}^{1/2}\frac{2t}{\sqrt{1+{t}^{2}}}dt$

Let $1+{t}^{2}=w$

$2tdt=dw$

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

When, $t	o 2$, then $w=5$

$\begin{gathered} {I}_{1} ={\int }_{5}^{5/4}\frac{1}{\sqrt{w}}dw \ \ =[2\sqrt{w}{]}_{5}^{5/4} \ \ =2[\sqrt{5}-\frac{\sqrt{5}}{2}]=\sqrt{5} \end{gathered}$

On substitute value of ${I}_{1}$ in Eq. (ii), we get

$I={\log }_{e}(\frac{\sqrt{2}(2+\sqrt{5}{)}^{2}}{\sqrt{1+\sqrt{5}}})-\frac{\sqrt{5}}{2}$

Solutions · 1