Question

integral from 0 to two ∫² 2t÷(t-3)²

Answer 1

yest i did but what happens to the 2t

Answer 2

i dont understand how to integrate

Answer 3

its easy, just takes some practice

Answer 4

u=t-3
du=dt
u+3=t
2u+6=2t
when t=0 u=-3
when t=2 u=-1
\[\int\limits_{-3}^{-1}2u+6/u^2du\]
\[\int\limits_{-3}^{-1}2/u+6/u^2du=2\ln|u|-6/u|[-1,-3]\]
2ln(1)+(-6/-1)-(2ln(3)+(-6/-3)=
0+6-2ln(3)-2
4-ln(9)=number but thats the purest answer

Answer 5

[-3,-1] not the opposite

Answer 6

\[\int\limits_0^2 \frac{2t}{(t-3)^2} \, dt=4-\text{Log}[9] \]