Question

steps please : integrate 1/(2x+3) from 0 to 3

Answer 1

Do a \(u=2x=3\) substitution.

Answer 2

i know that the integral of 1/(2x+3) is 1/2 ln(2x+3)+C but i need the steps to do from ) to 3

Answer 3

\(u = 2x+3\)

Answer 4

Did you learn about u substitution yet?

Answer 5

u substitution:

u = 2x+3
du = 2 dx

\[\rightarrow \frac{1}{2}\int\limits_{0}^{3} \frac{1}{u} du\]

Answer 6

yeah i know how to find the integral just forgot how to do from 0 to 3

Answer 7

is it like:
1/2 ln (9) - 1/2 ln(-3)?

Answer 8

oh you need to use Fundamental Thm of Calc
\[\int\limits_{a}^{b}f(x) dx = F(b) - F(a)\]
so
\[\frac{1}{2}[\ln (2*3+1) - \ln (2*0+1)]\]

Answer 9

Thanks :)

Answer 10

yw

Answer 11

wait shouldnt it be 1/2 (ln 2*3+3) and not 1?

Answer 12

yup, its +3 and not +1
that was probably just a typo

Answer 13

alright, was confused. Thanks!