We consider the functions $f(x) = x\mathrm{e}^{1-x^{2}}$ and $g(x) = \mathrm{e}^{1-x}$.

1. Find a primitive $F$ of the function $f$ on $\mathbb{R}$.
2. Deduce the value of $\int_{0}^{1} \left(\mathrm{e}^{1-x} - x\mathrm{e}^{1-x^{2}}\right) \mathrm{d}x$.
3. Give a graphical interpretation of this result.

Consider the function $f$ defined on $\mathbb{R}$ by $$f(x) = x\mathrm{e}^{x^2-3}$$ One of the antiderivatives $F$ of the function $f$ on $\mathbb{R}$ is defined by: a. $F(x) = 2x\mathrm{e}^{x^2-3}$ b. $F(x) = \left(2x^2+1\right)\mathrm{e}^{x^2-3}$ c. $F(x) = \frac{1}{2}x\mathrm{e}^{x^2-3}$ d. $F(x) = \frac{1}{2}\mathrm{e}^{x^2-3}$

Let
$$I = \int \frac { e ^ { x } } { e ^ { 4 x } + e ^ { 2 x } + 1 } d x , \quad J = \int \frac { e ^ { - x } } { e ^ { - 4 x } + e ^ { - 2 x } + 1 } d x$$
Then, for an arbitrary constant $C$, the value of $J - I$ equals
(A) $\frac { 1 } { 2 } \log \left( \frac { e ^ { 4 x } - e ^ { 2 x } + 1 } { e ^ { 4 x } + e ^ { 2 x } + 1 } \right) + C$
(B) $\frac { 1 } { 2 } \log \left( \frac { e ^ { 2 x } + e ^ { x } + 1 } { e ^ { 2 x } - e ^ { x } + 1 } \right) + C$
(C) $\frac { 1 } { 2 } \log \left( \frac { e ^ { 2 x } - e ^ { x } + 1 } { e ^ { 2 x } + e ^ { x } + 1 } \right) + C$
(D) $\frac { 1 } { 2 } \log \left( \frac { e ^ { 4 x } + e ^ { 2 x } + 1 } { e ^ { 4 x } - e ^ { 2 x } + 1 } \right) + C$

The integral $\int \frac { 2 x ^ { 3 } - 1 } { x ^ { 4 } + x } d x$ is equal to (here $C$ is a constant of integration)
(1) $\frac { 1 } { 2 } \ln \frac { | x ^ { 3 } + 1 | } { x ^ { 2 } } + C$
(2) $\frac { 1 } { 2 } \ln \frac { ( x ^ { 3 } + 1 ) ^ { 2 } } { | x ^ { 3 } | } + C$
(3) $\ln \frac { | x ^ { 3 } + 1 | } { x ^ { 2 } } + C$
(4) $\ln \frac { | x ^ { 3 } + 1 | } { x ^ { 3 } } + C$

If $\int \frac{\cos\theta}{5 + 7\sin\theta - 2\cos^2\theta}\,d\theta = A\log_e|B(\theta)| + C$, where $C$ is a constant of integration, then $\frac{B(\theta)}{A}$ can be:
(1) $\frac{2\sin\theta+1}{\sin\theta+3}$
(2) $\frac{2\sin\theta+1}{5(\sin\theta+3)}$
(3) $\frac{5(\sin\theta+3)}{2\sin\theta+1}$
(4) $\frac{5(2\sin\theta+1)}{\sin\theta+3}$