Question

\[\LARGE \int \frac{(5-v)dv}{-2(v^2 - v - 2)}\]

Answer 1

= -1/2 ∫ ( 5 - v) / ( v² -v -2)
Adjust the numerator a little bit:
= 1/4) ∫( 2v -1 ) / ( v² -v -2) dv - 9/4 ∫ dv/ ( v-2) ( v +1)

Answer 2

would you happen to know how to do this via partial fractions?

Answer 3

i think it's possible

Answer 4

Split them out by numerator:
= -5/2 ∫ dv/ ( v² -v -2) - ∫ v / ( v² -v -2)

Answer 5

-5/2?

Answer 6

Because when you split it the first part of numerator is 5

Answer 7

oh lol yeah of course

Answer 8

The second part: 1/2 ∫ v / ( v² -v -2)

Answer 9

this is where i do the P.F.?

Answer 10

Any method works fine :)

Answer 11

v^2 - v - 2 = (v-2)(v+1)

A(v+1) B(v-2) = v

when v = 2

A(3) = 2
A = 2/3

when v = -1
B(-3) = -1
B = 1/3

\[\frac{v}{v^2 - v - 2} \implies \int \frac{2}{3(v-2)} + \int \frac{1}{3(v+1)}?\]

Answer 12

Yep :)

Answer 13

okay thanks :D i think i can go from here now

Answer 14

I'm just back! Glad to help :)