Question

ok .. this integral was in my test today ... !!! x^3ln(1-x^(-4)) ... how would you solve it ? because i didn't =[

Answer 1

your teacher must hate you

Answer 2

Use integration by parts.

Answer 3

this is a set up for integration by parts, but really i think it will take several steps

Answer 4

i managed to get rid of the x^3 in the test .. but the i was stuck with the integral ln(1-1/x)

Answer 5

oh maybe not
\[u=\ln(1-x^{-4}), du = \frac{-4}{x-x^5}, dv = x^3, v = \frac{x^4}{4}\] maybe it is not so bad

Answer 6

second integral turns in to
\[\int\frac{x^3}{x^4-1}dx\] so a u -sub cures taht one

Answer 7

maybe we will see the chart method.

Answer 8

i was stuck on \[\int 2x^3 e^{2x}dx\]:) i couldnt parse it correctly on the test to save my life

Answer 9

e^(x^2) not e^(2x)

Answer 10

what is the chart method ? and how did you get to that integral ? can you show ?

Answer 11

amistre makes a chart, i am not sure how

Answer 12

i use crayons lol

Answer 13

i was pretty sure you were not allowed sharp objects...

Answer 14

:)

Answer 15

second integral is
\[\int v du\] so it is
\[\frac{x^4}{4}\times \frac{-4}{x-x^5}dx=-\int \frac{x^3}{1-x^4}dx\]

Answer 16

i think the problem may have been you were thinking of reducing the power on x^3, when instead you should focus on getting rid of the log

Answer 17

since the product rule is the basis for integration by parts; can it be adapted for multiple products?

\[\int fgh = something profound?\]

Answer 18

good question. will come back to it