Use the substitution $t = e^{x}$ to show that $\int e^{2x}\tan^{-1}\left(e^{x}\right)\,dx = \int t\tan^{-1}(t)\,dt$. Hence find the exact value of $\displaystyle \int_{0}^{1} e^{2x}\tan^{-1}\left(e^{x}\right)\,dx$.

1. Find the expression for $dx$ in terms of $t$ and $dt$.

Let $t = e^x$.
Given this substitution, differentiate with respect to $x$:
$\frac{dt}{dx} = e^x$
$dx = \frac{1}{e^x}\,dt = \frac{1}{t}\,dt$

2. Substitute $t$ and $dx$ into the integral and simplify.

Expressing $e^{2x}$ as $(e^x)^2 = t^2$, substitute all parts into the integral:
$\int e^{2x}\tan^{-1}(e^x)\,dx$
$= \int t^2 \tan^{-1}(t) \left(\frac{1}{t}\right)\,dt$
$= \int t\tan^{-1}(t)\,dt \quad \text{(Shown)}$

1. Determine the new limits of integration for the substitution $t = e^x$.

When $x = 0$, $t = e^0 = 1$.
When $x = 1$, $t = e^1 = e$.
Therefore,
$\int_{0}^{1} e^{2x}\tan^{-1}(e^{x})\,dx = \int_{1}^{e} t\tan^{-1}(t)\,dt$

2. Apply integration by parts.

We use integration by parts for $\int t\tan^{-1}(t)\,dt$.
Let $u = \tan^{-1}(t)$ and $\frac{dv}{dt} = t$.
Then $\frac{du}{dt} = \frac{1}{1+t^2}$ and $v = \frac{1}{2}t^2$.
$\int_{1}^{e} t\tan^{-1}(t)\,dt = \left[ \frac{1}{2}t^2\tan^{-1}(t) \right]_{1}^{e} - \int_{1}^{e} \frac{t^2}{2(1+t^2)}\,dt$

3. Simplify and integrate the rational function.

To integrate $\frac{t^2}{1+t^2}$, we rewrite the numerator:
$\int \frac{t^2}{1+t^2}\,dt = \int \frac{(t^2+1)-1}{1+t^2}\,dt = \int \left( 1 - \frac{1}{1+t^2} \right)\,dt$
$= t - \tan^{-1}(t)$
Substituting this back into our definite integral:
$\int_{1}^{e} t\tan^{-1}(t)\,dt = \left[ \frac{1}{2}t^2\tan^{-1}(t) - \frac{1}{2}\big(t - \tan^{-1}(t)\big) \right]_{1}^{e}$
$= \left[ \frac{1}{2}(t^2+1)\tan^{-1}(t) - \frac{1}{2}t \right]_{1}^{e}$

4. Substitute the limits and evaluate to find the exact value.

Substitute the limits $t = e$ and $t = 1$:
$\left( \frac{1}{2}(e^2+1)\tan^{-1}(e) - \frac{1}{2}e \right) - \left( \frac{1}{2}(1^2+1)\tan^{-1}(1) - \frac{1}{2}(1) \right)$
Since $\tan^{-1}(1) = \frac{\pi}{4}$:
$= \frac{1}{2}(e^2+1)\tan^{-1}(e) - \frac{1}{2}e - \left( \frac{\pi}{4} - \frac{1}{2} \right)$
$= \frac{1}{2}(e^2+1)\tan^{-1}(e) - \frac{1}{2}e - \frac{\pi}{4} + \frac{1}{2}$