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Quick C3 integration

Was hoping someone could help on this please, integrating with limits 1 to 9
π∫363x−2\large \pi \int \frac{36}{3x-2}

I pulled out the 36 from the numerator , and made it:
36π∫13x−2\large 36\pi \int \frac{1}{3x-2}

36πln(3x−2)\large 36\pi ln(3x-2)

After subbing in the limits I get the answer wrong as the answer given involves 12 pi ?

I got 36Pi ln25 , answer is 12Pi ln25 , I thought my method would usually work for these.
(edited 11 years ago)
Reply 1
Avatar for RajPopat94
RajPopat94
When you're integrating lnf(x), what's the rule?
The format has to be f'(x)/f(x). So you have to take out 12, as then you're left with 3 at the top (the bottom differentiates to 3). You should be fine from there on out.
notice you need a third of it
Reply 4
Avatar for Tulian
Tulian
OP
Original post by RajPopat94
When you're integrating lnf(x), what's the rule?


If it's 1/something, then ln(something) right ? , but I took out the 36 as a constant so that shouldn't matter ?
Reply 5
Avatar for raheem94
raheem94
13
Original post by Tulian
If it's 1/something, then ln(something) right ? , but I took out the 36 as a constant so that shouldn't matter ?


∫cax+b dx=c∫1ax+b dx=c×1aln∣ax+b∣+C \displaystyle \int \frac{c}{ax+b} \ dx = c \int \frac{1}{ax+b} \ dx = c \times \frac1{a} ln|ax+b| + C
Reply 6
Avatar for Tulian
Tulian
OP
I totally forgot about the 1/a part after taking the 36 out.

Thanks everyone

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