Solve the integral \[\int {\dfrac{{1 + v}}{{1 - 2v - {v^2}}}} \].

Hint: In this problem, we need to solve the given integral function by using differentiation and integration. The term integral can refer to a number of different concepts in mathematics. In calculus, an integral is a mathematical object that can be interpreted as an area or a generalization of area. Integrals, together with derivatives, are the fundamental objects of calculus. Integration is the algebraic method of finding the integral for a function at any point on the graph. The integral is usually called the antiderivative, because integrating is the reverse process of differentiating.

Complete step by step solution:
In the given problem,
The integral function is \[\int {\dfrac{{1 + v}}{{1 - 2v - {v^2}}}} dv\]
Let the denominator function as \[z = 1 - 2v - {v^2}\]
\[z = 1 - 2v - {v^2}\]
By differentiating the function,\[z\] with respect to \[v\], we get
\[\
 dz = ( - 2 - 2v)dv \\
 dz = - 2(1 + v)dv \\
\ \]
To simplify, we get
\[(1 + v)dv = - \dfrac{1}{2}dz\]
Where, \[z = 1 - 2v - {v^2}\]
By substitute the value in the given equation, so we can get
\[
 \int {\dfrac{{1 + v}}{{1 - 2v - {v^2}}}} = \int {\dfrac{{(1 + v)}}{{1 - 2v - {v^2}}}dv} \\
 \int {\dfrac{{(1 + v)}}{{1 - 2v - {v^2}}}dv} = \int {\dfrac{{ - \dfrac{1}{2}dz}}{z}} \\
\]
Now, we have
\[ = - \dfrac{1}{2}\int {\dfrac{1}{z}dz} \]
On comparing the integration formula \[\int {\dfrac{1}{x}dx = {\rm I}n\left| x \right| + C} \] in further step, we get
\[ = - \dfrac{1}{2}{\rm I}n\left| z \right| + C\].
We know that, \[z = 1 - 2v - {v^2}\]by substitute in the above equation, we get
\[ = - \dfrac{1}{2}{\rm I}n\left| {1 - 2v - {v^2}} \right| + C\]
 Here, \[C\] is the integral constant.
Therefore, the final answer is \[\int {\dfrac{{1 + v}}{{1 - 2v - {v^2}}}} = - \dfrac{1}{2}{\rm I}n\left| {1 - 2v - {v^2}} \right| + C\].

Note:
Here, we use the integration and differentiation in the given function. Integral is defined as the number of different concepts in mathematics. Integrals, together with derivatives, are the fundamental objects of calculus. Integration is the algebraic method of finding the integral for a given function, we use this formula \[\int {\dfrac{1}{x}dx = {\rm I}n\left| x \right| + C} \] to get the solution. In calculus, an integral is a mathematical object that can be interpreted as an area or a generalization of area.