Question

how to integrate \[\int \frac{8}{4v^2 +1}dv\] lol i forgot how to do these

Answer 1

Can try taking the constant, 8, out, and then do a backwards-chain rule operation that will leave you with a natural log expression..

Answer 2

\[\int \frac{8}{4v^2 +1}dv\]
\[=\int \frac{8}{(2v)^2 +1}dv\]
Let\[2v=\tan(\theta)\]...

Answer 3

*facepalm* tangent of course

Answer 4

:)

Answer 5

I had a feeling a trig-sub would work. Whenever you see a sum-of-squares, think trig-sub!

Answer 6

is it 1/2 arctan 2v?

Answer 7

close

Answer 8

you forgot your 8 i believe

Answer 9

\[4\,\tan^{-1}(2v)+c\]

Answer 10

8?

Answer 11

oh i see

Answer 12

\[\int \frac{8}{4v^2 +1}dv=8\int \frac{1}{4v^2 +1}dv\]

\[=8\frac{1}{2}\arctan(2v)+c=4\arctan(2v)+c\]