使用换元 $t = e^{x}$ 证明：$\int e^{2x}\,\tan^{-1}\bigl(e^{x}\bigr)\,dx = \int t\,\tan^{-1}(t)\,dt.$ 由此求出 $\displaystyle \int_{0}^{1} e^{2x}\,\tan^{-1}\bigl(e^{x}\bigr)\,dx$ 的确切值。

1. Differentiate the substitution equation to find the differential relationship between $x$ and $t$.

$t = e^{x}$
$\frac{dt}{dx} = e^{x}$
$dt = e^{x}\,dx$

2. Rewrite the integral to isolate the differential $e^x dx$ and apply the substitution.

$\int e^{2x}\,\tan^{-1}(e^{x})\,dx = \int e^{x}\,\tan^{-1}(e^{x}) \cdot (e^{x}\,dx)$
Substituting $t = e^{x}$ and $dt = e^{x}\,dx$:
$= \int t\,\tan^{-1}(t)\,dt \quad \text{(shown)}$

1. Change the limits of integration according to the substitution $t = e^x$.

When $x = 0$, $t = e^0 = 1$.
When $x = 1$, $t = e^1 = e$.
Using the result from part (a):
$\int_{0}^{1} e^{2x}\tan^{-1}(e^x)\,dx = \int_{1}^{e} t\tan^{-1}(t)\,dt$

2. Apply integration by parts to the new integral.

Let $u = \tan^{-1}(t)$ and $\frac{dv}{dt} = t$.
Then $\frac{du}{dt} = \frac{1}{1+t^2}$ and $v = \frac{1}{2}t^2$.
$\int t\tan^{-1}(t)\,dt = \frac{1}{2}t^2\tan^{-1}(t) - \frac{1}{2} \int \frac{t^2}{1+t^2}\,dt$

3. Integrate the remaining rational function by rewriting the numerator.

Notice that $\frac{t^2}{1+t^2} = \frac{(t^2+1)-1}{1+t^2} = 1 - \frac{1}{1+t^2}$.
$\int \frac{t^2}{1+t^2}\,dt = \int \left( 1 - \frac{1}{1+t^2} \right)\,dt = t - \tan^{-1}(t)$

4. Evaluate the definite integral using the upper and lower limits.

$\int_{1}^{e} t\tan^{-1}(t)\,dt = \left[ \frac{1}{2}t^2\tan^{-1}(t) - \frac{1}{2}\bigl(t - \tan^{-1}(t)\bigr) \right]_{1}^{e}$
$= \left[ \frac{1}{2}(t^2+1)\tan^{-1}(t) - \frac{1}{2}t \right]_{1}^{e}$
Substituting the upper and lower limits:
$= \left( \frac{1}{2}(e^2+1)\tan^{-1}(e) - \frac{1}{2}e \right) - \left( \frac{1}{2}(1^2+1)\tan^{-1}(1) - \frac{1}{2}(1) \right)$
Since $\tan^{-1}(1) = \frac{\pi}{4}$:
$= \frac{1}{2}(e^2+1)\tan^{-1}(e) - \frac{1}{2}e - \left( 1 \cdot \frac{\pi}{4} - \frac{1}{2} \right)$
$= \frac{1}{2}(e^2+1)\tan^{-1}(e) - \frac{1}{2}e - \frac{\pi}{4} + \frac{1}{2}$