Question

Evaluate the definite integrals.
\[\int\limits_{0}^{4} (\frac{ 3 }{ 2x+1 })dx\]

Answer 1

u=2x+1
du=2dx
??

Answer 2

thats a good start

Answer 3

New limit
x=0, u=1
x=4, u=9

Answer 4

So it would like this?..
\[\int\limits_{1}^{9}\frac{ 3 }{ u }du\]
??

Answer 5

not quite, there is a mistake with differential

du=2dx
dx = ?

Answer 6

dx= 1/2 du ?

Answer 7

Yes

Answer 8

Oh so it would be:
\[\int\limits_{1}^{9}\frac{ 3 }{ 2u }du\]

Answer 9

Then I got 3/2ln9 as the final answer..?

Answer 10

looks good!

Answer 11

Is it possible to reduce that even more?

Answer 12

use the log property
\[\large n\,\log x ~~=~~\log x^n \]

Answer 13

I'm not sure how to use it correctly...

Answer 14

\[\large \frac{3}{2}\,\ln(9) ~~=~~\ln (9^{\frac{3}{2}}) \]

Answer 15

Then what happens?...

Answer 16

\[\large \frac{3}{2}\,\ln(9) ~~=~~\ln (9^{\frac{3}{2}}) =\ln({\sqrt{9~}}^3)\]

Answer 17

Oh I see! It would be 3ln3!

Answer 18

Yep!

Answer 19

Thank you! :)

Answer 20

yw!