Animated Solution for Mathematics - Definite Integration: The value of the integral ${\int }_{-\ln 2}^{\ln 2}{e}^x(\ln (e^x+\sqrt{1+e^{2x}}))dx$ is equal to

The given integral is:

At first glance, this expression appears to be a daunting, tangled mess of exponentials and logarithms. However, in the context of JEE Advanced, appearances are often deceiving. The key to unlocking this problem lies in recognizing the hidden symmetry.

Notice how $e^x$ appears both as a factor and inside the logarithm. This is a classic invitation to use the substitution method. By setting $t=e^x$, we transform the variable $x$ into $t$, and the differential $e^{x}dx$ becomes $dt$.

When we perform the substitution $t=e^x$, we must also update our limits of integration. When $x=-\ln 2$, $t=e^{-\ln 2}=\frac{1}{2}$. When $x=\ln 2$, $t=e^{\ln 2}=2$.

Our integral now becomes:

This is a much cleaner, more elegant form. We are now looking for the area under the curve of the natural log of $t$ plus the square root of $1+t^2$.

To tackle the integral of a lone logarithmic function, we turn to our most powerful tool: Integration by Parts. We set $u=\ln (t+\sqrt{1+t^2})$ and $dv=dt$, which implies $v=t$.

The magic happens when we calculate $du$. Using the chain rule, the derivative of $\ln (t+\sqrt{1+t^2})$ is:

Simplifying this, we get:

The complexity vanishes!

Applying the Integration by Parts formula, $I=[uv]-\int vdu$, we obtain:

The second integral is a standard form, resulting in $\sqrt{1+t^2}$. Combining everything, we have:

Evaluating at the upper limit $t=2$ and the lower limit $t=\frac{1}{2}$, we calculate:

Simplifying the expression, the final result is:

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