Question

determine the integral: 1/square root of 3x - 7

Answer 1

need a 2 for the bottom and a 3 for the top; multiply by 6/6

Answer 2

\[\int\frac{2}{2}*\frac{3}{3}\frac{1}{\sqrt{3x-7}}\]
\[\frac{2}{3}\int\frac{3}{2\sqrt{3x-7}}\]

Answer 3

do you see why?

Answer 4

\[\int\limits \frac{dx}{\sqrt{3x-7}}\]
Let:
\[u=\sqrt{3x-7} \rightarrow u^2=3x-7\]
Then:
\[2 u du=3dx \rightarrow \frac{2}{3}udu=dx\]
Pluggin into the integral it becomes:
\[\frac{2}{3} \int\limits \frac{u du}{u}=\frac{2}{3} \int\limits du=\frac{2}{3}u+C=\frac{2}{3}\sqrt{3x-7}+C\]

Answer 5

lol .... you used a "u" :)

Answer 6

xP

Answer 7

You can also do u=3x-7. I just wasn't paying attention and did it the harder way.

Answer 8

i just multiplied by 1 ;)

Answer 9

I mean, yeah, but either way some people can see it off that :P

Answer 10

I'm not doubting your method amistre :P Even though I'M STILL NOT A MOD :'(

Answer 11

\[\color{purple}{\small\text{maybe tomorrow ?}}\]

Answer 12

Please :D

Answer 13

the position comes with a lantern and a ring; and a box of hohos

Answer 14

lantern ring and hohos sold seperately

Answer 15

I would be happy to accept the responsibility :)

Answer 16

when everyones a mod; then noones a mod ;)

Answer 17

But but butttttt I do alot on here :D I've only been here like 2 weeks and I have like 240 medals or something O:

Answer 18

ive been here since mid march i think ....

Answer 19

Week, feels like a year...