Question 1

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

Question 2

Answer:

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

Please log in or register to add a comment.

Please log in or register to answer this question.