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Sagot :
Answer:
[tex]\int\limits {\frac{V-1}{1+2V+V^{2} } } \, dv[/tex] [tex]= log V + (\frac{2 }{1+v} )+C[/tex]
Step-by-step explanation:
Step(i):-
Given
[tex]\int\limits {\frac{V-1}{1+2V+V^{2} } } \, dv[/tex]
= [tex]\int\limits {\frac{V-2+1}{1+2V+V^{2} } } \, dv[/tex]
[tex]= \int\limits {\frac{V+1}{(1+V)^{2} } } \, dv+\int\limits {\frac{-2}{(1+V)^{2} } } \, dv[/tex]
= [tex]= \int\limits {\frac{1}{(1+V) } } \, dv+\int\limits {\frac{-2}{(1+V)^{2} } } \, dv[/tex]
Step(ii):-
By using integration formula
[tex]\int\limits {\frac{1}{x} } \, dx =logx +C[/tex]
[tex]\int\limits {x^{n} } \, dx = \frac{x^{n+1} }{n+1} +C[/tex]
[tex]= \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[/tex]
[tex]= log V + (\frac{-2(1+v)^{-1} }{-1} )+C[/tex]
[tex]= log V + (\frac{2 }{1+v} )+C[/tex]
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