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Integration help

Hi I was wondering if you had to integrate 3x squared / 1-2x cubed how would you do it?

I known that it would be using ln but so it would be ln(1-2x cubed) but what happens to the top part of the fraction when integrating?
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Avatar for TeeEm
TeeEm
19
substitution or partial fractions
if it is

3x2
----------
1 - x3

then the top is nearly the derivative of the bottom so ln is the way to go...
Original post by the bear
if it is

3x2
----------
1 - x3

then the top is nearly the derivative of the bottom so ln is the way to go...


Hi, I know that it would be ln(1-x cubed) but what would happen to the 3x2
Original post by Teddysmith123
Hi, I know that it would be ln(1-x cubed) but what would happen to the 3x2


if you have to integrate

f'(x)
------
f(x)

the answer is just ln|f(x)| + c

the top does not reappear.

however in this case the top is not the exact derivative of the bottom...
Original post by Teddysmith123
Hi, I know that it would be ln(1-x cubed) but what would happen to the 3x2


It wouldn't be ln⁡∣1−x3∣\ln|1-x^3| though that is very close


Differentiate ln⁡∣1−x3∣\ln|1-x^3| and the answer to your question should be obvious

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