Let
So (1) becomes:
You can check the answer here to make sure.
int (0.8(3 + 2 ln(u))^3)/u - Wolfram|Alpha
Could you explain where I went wrong with this one?
28-1-16
(int) 0.8((3 + 2 ln(u))^3) du/u
Attempt:
u = 3 + 2 ln u
du = 1 / u
Therefore answer should be:
(.8 [3 + 2 ln u]^3) / 4
the ti-89 comes up with another answer when this integration is performed, which is long and bizarre.
Let
So (1) becomes:
You can check the answer here to make sure.
int (0.8(3 + 2 ln(u))^3)/u - Wolfram|Alpha
You are almost on the right track.
Let
So plugging in and we get:
So,
You can check the answer by taking the derivative of which
is equal to
derivative 0.8((3 + 2 ln(u))^4}/8) + C - Wolfram|Alpha
Hope it answers your question.
Integration is the opposite of differentiation hence the other name anti-derivative.
So by integrating what you'll get is when differentiated will be what you're integrating.
So:
you're integrating and you'll get
When you differentiate you'll get what you're integrating. So when you integrate you've to rearrange variables and constants in such a way that when you differentiate the rearrangement you'll end up with what you're integrating.
So if we differentiate we get:
Let which is what you're integrating.
And that's why
what confuses me is according to the general power rule, you add 1 to the exponent found on the integral, which is 3, so ends up being 4 on the bottom. While I understand it differentiates properly (post #6), seems like something other than general power rule was used to come up with a denominator of 8.