Question

Not sure how they got to this answer in my sol. manual.
The integral of 2/(v^2-v) dv = 2ln[(v-1)/v]

Answer 1

\[\int\limits\limits_{}^{} 2/(v^2-v) dv = 2\ln ((v-1)/v)\]

Answer 2

take v common in the denominator, ull get 2/v(v-1) dv.
Now you can take partial fractions. Understood?

Answer 3

partial fractions
v^2 -v = v(v-1)
A/v + B/v-1 = A(v-1) + Bv / v(v-1) = 2/v(v-1)
(A+B)v - A = 2
-A = 2 --> A = -2
A+B = 0 --> B=2
integral -2/v + 2/v-1
-2ln(v) +2ln(v-1) +C
= 2ln(v-1/v) +C

Answer 4

you need to present v^2-v=v(v-1)
after that:
A/v + B/(v-1)=...
find A & B
now your integral could be presented as:
\[=\int\limits_{ }^{}-2dv/v + \int\limits_{ }^{}2dv/(v-1)=...\]
i think you can take it from here...?

Answer 5

Yes! Thank you!