Question

integral of (3v/(1-4v^2))

Answer 1

Just let u be the denominator for your u-substitution. u = 1-4v^2, so du = -8vdx and dx = -du/8v

Answer 2

\[x(dv/dx)=(1-4v^2/3v) \]

Answer 3

i have to solve the equation

Answer 4

3v is dividing all of 1-4v^2?

Answer 5

yea

Answer 6

Im guessing it is considering your first question. Then yeah, looked like you had the write idea, separation of variables. So we have:
\[\int\limits_{}^{}\frac{ 3v }{ 1-4v ^{2} }dv=\int\limits_{}^{}\frac{ 1 }{ x }dx\]So as I aning mentioned before, you can let u be 1-4v^2. meaning du = -8vdv and dv = -du/8v. The right integral is just an immediate integration into ln. So all of this gives us:
\[3\int\limits_{}^{}\frac{ }{ }\frac{ v }{ u }*\frac{ -du }{ 8v }=\int\limits_{}^{}\frac{ 1 }{ u }du\]
\[-\frac{ 3 }{ 8 }\ln|1-4v ^{2}|=\ln|x|+C\]

Would you know how to take it from there?

Answer 7

yea thanks so much!!!

Answer 8

Yep, np ^_^