
Jacob C. answered 04/23/21
College student just trying to help out
integral (6t)/(1+2t) from 0 to 3.
6t/1+2t = (6t+3-3)/(2t+1) =
(6t+3)/(2t+1) + (-3)/(2t+1) = 3(2t+1)/(2t+1) + (-3)/(2t+1) =
3 + (-3)/(2t+1) now apply the integral and limits, should be straight forward. Turns into
3t - (3/2)ln(2t+1) from 0 to 3
9 - (3/2)ln(7) - ((-3/2)ln(1))
final answer 9-(3/2)ln(7)
Hope this helps.

Sreeram K.
correct! thanks alot!04/24/21
Jacob C.
As a decimal it would be 6.08104/24/21