Question
Question: Solve the integral \[\int {\dfrac{{1 + v}}{{1 - 2v - {v^2}}}} \]....
Solve the integral ∫1−2v−v21+v.
Solution
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 ∫1−2v−v21+vdv
Let the denominator function as z=1−2v−v2
z=1−2v−v2
By differentiating the function,z with respect to v, we 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}} \\