Question

Integrate V-1 divided by 1+2V+V^2

Answer 1

`\int\limits {(V-1)/(1+2V+V^(2) ) } \, dv`
`= log V + ((2 )/(1+v) )+C`

Explanation:

Step(i):-

Given

`\int\limits {(V-1)/(1+2V+V^(2) ) } \, dv`

=
`\int\limits {(V-2+1)/(1+2V+V^(2) ) } \, dv`

`= \int\limits {(V+1)/((1+V)^(2) ) } \, dv+\int\limits {(-2)/((1+V)^(2) ) } \, dv`

=
`= \int\limits {(1)/((1+V) ) } \, dv+\int\limits {(-2)/((1+V)^(2) ) } \, dv`

Step(ii):-

By using integration formula

`\int\limits {(1)/(x) } \, dx =logx +C`

`\int\limits {x^(n) } \, dx = (x^(n+1) )/(n+1) +C`

`= \int\limits {(1)/((1+V) ) } \, dv+\int\limits {(-2)/((1+V)^(2) ) } \, dv\\= log V + ((-2(1+v)^(-2+1) )/(-2+1) )+C`

`= log V + ((-2(1+v)^(-1) )/(-1) )+C`

`= log V + ((2 )/(1+v) )+C`