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 :

Question

Accepted Answer

Answer: B. \log _{e}\left(\frac{\sqrt{2}(2+\sqrt{5})^{2}}{\sqrt{1+\sqrt{5}}}ight)-\frac{\sqrt{5}}{2} \int\limits_{-\log _{e} 2}^{\log _{e} 2} e^{x}\left(\log _{e}\left(e^{x}+\sqrt{1+e^{2 x}}ight)ight) d x Let e^x=t \Rightarrow e^x d x=d t When, x ightarrow-\log _e 2, then t ightarrow \frac{1}{2} When, x ightarrow \log _e 2, then t ightarrow 2 I=\int\limits_{\frac{1}{2}}^2\left[\log _e\left(t+\sqrt{1+t^2}ight)ight] 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}ight)ight]_{1 / 2}^2-\int_{1 / 2}^2 \frac{t}{t+\sqrt{1+t^2}}\left(1+\frac{2 t}{2 \sqrt{1+t^2}}ight) d t \\ & =2 \log _e(2+\sqrt{5})-\frac{1}{2} \log _e\left(\frac{1+\sqrt{5}}{2}ight)-\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}ight)^{1 / 2}}ight)-\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 ightarrow \frac{1}{2}, then w=\frac{5}{4} When, t ightarrow 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}ight]=\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}}}ight)-\frac{\sqrt{5}}{2}

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 :

Answer

Explanation

More JEE Math questions