Question

Please could someone let me know if my answer is correct?
The problem is the integral of1/(3x-2)dx with the upper limit 1 and the lower limit 0. I have the answer of: -ln(2)/(3). s this answer correct? Thank you!

Answer 1

question: is the equation like this?

\[\int\limits_{}^{}\frac{ dx }{ 3x - 2 }\]

if so then the answer should be:

\[\frac{ 1 }{ 3 } \ln 3x - 2\]

because the derivative of 3x is 3dx.

so you need to put 1/3 outside. also:

dx/ x+1 = ln (x+1)

only if the numerator is the derivative of the denominator.

Answer 2

Are you a FLVS student or CCA?

Answer 3

ok, so I get the part about 1/3 on outside.
Because also worked it out to be the answer of: 1/3 [-ln2].

My limits changed to upper 1and lower -2
=1/3[ ln |u|]
=1/3[ln1-ln2]
=1/3[-ln2]


w