Hello, I just finished the first problem I posted and was wondering if I could get it double checked? Thanks in advance!
Integral (0,2) of [(2/(x+1) + (3/(x-4)]dx
Integral (0,2) 2/(x+1) dx + Integral (0,2) 3/(x-4) dx
Let u = x+1; Let u = x-4;
du/dx = 1 du/dx = 1
du=dx du=dx
Integral (0,2) 2/u du + Integral (0,2) 3/u du
2*Integral 1/u du + 3*Integral 1/u du
2lnu + 3lnu
2ln(x+1) + 3ln(x-4)
Plugging in the intervals:
[2ln(2+1) + 3ln(2-4)] - [2ln(1) + 3ln(-4)]
[ln(3)^2 + ln(-2)^3] - [ln(1)^2 + ln(-4)^3]
[ln(9) + ln(-8)] - [ln(1) + ln(-64)]
ln(-72) - ln(-64)
ln(-72/-64)
ln (9/8) = 0.0511525224