Looking for answers? Westonci.ca is your go-to Q&A platform, offering quick, trustworthy responses from a community of experts. Explore a wealth of knowledge from professionals across different disciplines on our comprehensive platform. Connect with a community of professionals ready to provide precise solutions to your questions quickly and accurately.
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]
Thank you for choosing our service. We're dedicated to providing the best answers for all your questions. Visit us again. We hope our answers were useful. Return anytime for more information and answers to any other questions you have. Thank you for choosing Westonci.ca as your information source. We look forward to your next visit.
