Question

Integration help?

Answer 1

u=ln 3x^2
du=6x/3x^2

Answer 2

On the right track, but before doing that simplify ln 3x^2

Answer 3

2x/x^2

Answer 4

2/x

Answer 5

good. du = 2/x . Do you see anything like du in the integral?

Answer 6

just the 1/x
multiply by 1/2?

Answer 7

good. So now put in all the substitutions....

Answer 8

\[\frac{ 1 }{ 2 } \int\limits_{}^{} u\]

right?

Answer 9

close. We have to be careful about the ln 3x^2....

Answer 10

\[\frac{ (\ln 3x^2)^2 }{ 4 }\]

Answer 11

\[\frac{ 1 }{ 2 } \int\limits_{}^{} \ln u\]?

Answer 12

im not sure

Answer 13

Sorry, I had to work it out myself. Your first attempt was good(remember the +C).

Answer 14

np

Answer 15

@legomyego180
Are you ready to continue?

Answer 16

Actually, you have got the correct answer.
\((1/2)\int udu\)=(1/2)u^2/2=u^2/4=ln(3x^2)/4+C
You started with the good substitution.

Answer 17

hmm if you notice I think we can set u to ln(3x^{2})

\[\int\limits \ln(3x^{2})*(\frac{ 1 }{ x })dx \]

Answer 18

Sorry guys, I've figured it out. All of the help is appreciated

Answer 19

\[u = \ln(3x^{2}) s = 3x^{2} =>\frac{ 1 }{ 3x^{2} }*(\frac{ ds }{ dx }) => \frac{ 6x }{ 3x^{2} }\]

\[du = (\frac{ 2 }{ x })(\frac{ 1 }{ 2 }) = \frac{ 1 }{ x }~dx \]

\[\frac{ 1 }{ 2 } \int\limits u~du\]

\[\frac{ 1 }{ 2 } \frac{ u^{2} }{ 2 }+C\]

Answer 20

which checks out to:

\[\frac{ (\ln(3x^{2}))^{2} }{ 4 }+C\]

had to do this myself too :)