Question

integrate (ln x)^6/x

Answer 1

Lana, maybe I should post my calculus project on here. Maybe then you'll help me

Answer 2

let u=lnx
du=1/x
so the integration becomes

\[\int\limits{u^6du} = \frac{u^7}{7}\]
substitute u=lnx back\[\int\limits{\frac{(lnx)^6}{x}dx} = \frac{(lnx)^7}{7}+C\]

Answer 3

your awesome!

Answer 4

what about this one?

Answer 5

dy/dx=16x(4x^2-1)^3 passing through (1,3)

Answer 6

whats the question? u want to integrate it?

Answer 7

yes

Answer 8

well i have to find the equation

Answer 9

\[y=\int\limits{16x(4x^2-1)^3dx}\]

Answer 10

let u= 4x^2-1

Answer 11

so y becomes\[\frac{1}{2} \times \frac{ (4x^2-1)^5}{5}+C\]

Answer 12

hmm that isn't an option

Answer 13

what are the options maybe i mad a lil mistake wait

Answer 14

the integration is right, u just have t plug in the point that it is passing through

Answer 15

the options only have powers of 4 and there is no fraction with 4x^2-1

Answer 16

oh now i know what the mistake is lol

Answer 17

let u=4x^2-1
du=8xdx
\[y=\int\limits{2u^3du}\]\[=\frac{1}{2}\frac{u^4}{4}\]

Answer 18

i mean\[\frac{1}{2}u^4\]