Question

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

Answer 1

Answer:

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

Step-by-step explanation:

Step(i):-

Given  

                $\int\limits {\frac{V-1}{1+2V+V^{2} } } \, dv$

           =   $\int\limits {\frac{V-2+1}{1+2V+V^{2} } } \, dv$

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

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

Step(ii):-

By using integration formula

$\int\limits {\frac{1}{x} } \, dx =logx +C$

$\int\limits {x^{n} } \, dx = \frac{x^{n+1} }{n+1} +C$

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

   $= log V + (\frac{-2(1+v)^{-1} }{-1} )+C$

  $= log V + (\frac{2 }{1+v} )+C$