Differentiate $\mathrm{e}^{x^2+2x}$ with respect to $x$ and hence find the exact value of $\int_{0}^{1} (x+1)\mathrm{e}^{x^2+2x} \; \mathrm{d}x$. Using the substitution $x = \sin t$, where $0 < t < \frac{\pi}{2}$, find $\int \frac{1}{(1-x^2)^{\frac{3}{2}}} \; \mathrm{d}x$ in terms of $x$.

1. Differentiate $\mathrm{e}^{x^2+2x}$ with respect to $x$ using the chain rule.

$\frac{\mathrm{d}}{\mathrm{d}x} (\mathrm{e}^{x^2+2x}) = (2x+2)\mathrm{e}^{x^2+2x}$
$= 2(x+1)\mathrm{e}^{x^2+2x}$

2. Relate the given integral to the previously found derivative (reverse differentiation).

Since $\frac{\mathrm{d}}{\mathrm{d}x} (\mathrm{e}^{x^2+2x}) = 2(x+1)\mathrm{e}^{x^2+2x}$, we can integrate both sides:
$\int 2(x+1)\mathrm{e}^{x^2+2x} \; \mathrm{d}x = \mathrm{e}^{x^2+2x}$
$\int (x+1)\mathrm{e}^{x^2+2x} \; \mathrm{d}x = \frac{1}{2}\mathrm{e}^{x^2+2x}$

3. Evaluate the definite integral by substituting the limits $x = 0$ and $x = 1$.

$\int_{0}^{1} (x+1)\mathrm{e}^{x^2+2x} \; \mathrm{d}x = \left[ \frac{1}{2}\mathrm{e}^{x^2+2x} \right]_0^1$
$= \frac{1}{2}(\mathrm{e}^{1^2+2(1)} - \mathrm{e}^{0^2+2(0)})$
$= \frac{1}{2}(\mathrm{e}^3 - \mathrm{e}^0)$
$= \frac{1}{2}(\mathrm{e}^3 - 1)$

1. Differentiate the substitution equation to express $\mathrm{d}x$ in terms of $\mathrm{d}t$.

Given the substitution $x = \sin t$, we differentiate with respect to $t$:
$\frac{\mathrm{d}x}{\mathrm{d}t} = \cos t \implies \mathrm{d}x = \cos t \; \mathrm{d}t$

2. Substitute $x$ and $\mathrm{d}x$ into the integral and use a trigonometric identity.

Substitute into the integral and apply the Pythagorean identity $1 - \sin^2 t = \cos^2 t$:
$\begin{aligned} \int \frac{1}{(1-x^2)^{\frac{3}{2}}} \; \mathrm{d}x &= \int \frac{1}{(1-\sin^2 t)^{\frac{3}{2}}} (\cos t) \; \mathrm{d}t \\ &= \int \frac{\cos t}{(\cos^2 t)^{\frac{3}{2}}} \; \mathrm{d}t \end{aligned}$

3. Simplify the integrand and perform the integration.

Since $0 < t < \frac{\pi}{2}$, $\cos t$ is positive, so $(\cos^2 t)^{\frac{3}{2}} = \cos^3 t$. Simplifying and integrating gives:
$\begin{aligned} \int \frac{\cos t}{\cos^3 t} \; \mathrm{d}t &= \int \frac{1}{\cos^2 t} \; \mathrm{d}t \\ &= \int \sec^2 t \; \mathrm{d}t \\ &= \tan t + C \end{aligned}$

4. Convert the integrated expression back into terms of $x$.

Since $x = \sin t$ and $0 < t < \frac{\pi}{2}$, we can represent this using a right-angled triangle with angle $t$, opposite side $x$, and hypotenuse $1$.
The adjacent side is $\sqrt{1^2 - x^2} = \sqrt{1 - x^2}$.
Therefore, $\tan t = \frac{\text{opposite}}{\text{adjacent}} = \frac{x}{\sqrt{1 - x^2}}$.

The integral evaluates to:
$\frac{x}{\sqrt{1 - x^2}} + C$