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

(D) $\log _{e}\left(\frac{\sqrt{2}(2+\sqrt{5})^{2}}{\sqrt{1+\sqrt{5}}}\right)-\frac{\sqrt{5}}{2}$

(A) $\log _{e}\left(\frac{2(2+\sqrt{5})}{\sqrt{1+\sqrt{5}}}\right)-\frac{\sqrt{5}}{2}$

(B) $\log _{e}\left(\frac{\sqrt{2}(3-\sqrt{5})^{2}}{\sqrt{1+\sqrt{5}}}\right)+\frac{\sqrt{5}}{2}$

(C) $\log _{e}\left(\frac{(2+\sqrt{5})^{2}}{\sqrt{1+\sqrt{5}}}\right)+\frac{\sqrt{5}}{2}$

JEE Main · 2023 · 11 Apr 2023 · Shift 1 · Definite Integrals

The value of the integral

\int_{- l o g_{e} 2}^{l o g_{e} 2} e^{x} (lo g_{e} (e^{x} + 1 + e^{2 x})) d x

is equal to

Answer: (D)

Substitute

e^{x} = t

, so

e^{x} d x = d t

. The limits become

t = 1/2

t = 2

. The integral becomes

I = \int_{1/2}^{2} ln (t + 1 + t^{2}) d t

. Apply integration by parts with

u = ln (t + 1 + t^{2})

and

d v = d t

. This gives

[t ln (t + 1 + t^{2})]_{1/2}^{2} - \int_{1/2}^{2} \frac{t}{1 + t ^{2}} d t

. Evaluating:

2 ln (2 + 5) - \frac{1}{2} ln (\frac{1 + 5}{2}) - \frac{5}{2} = ln (\frac{( 2 + 5 ) ^{2}}{( 1 + 5 ) /2}) - \frac{5}{2} = ln (\frac{2 ( 2 + 5 ) ^{2}}{1 + 5}) - \frac{5}{2}

JEE Main · 2023 · 11 Apr 2023 · Shift 1 · Single correct option