Question

Integrate ln(x)/x using the general substitution rule... I have made u=ln(x) and du= dx/x...

Answer 1

so you got Integral of U du what is that?

Answer 2

Yah it looks like you're on the right track Finn. Just confused about how to setup the U integral?

Answer 3

yes, exactly... Zepdrix, we're going to become friends tonight I feel... ok, let's do this

Answer 4

lol XD

Answer 5

i think i know what ur missing
maybe you forgot that "Integrate sign" Ln(x)/x * dx so you can replace 1/x * dx with du as du = 1/x * dx

Answer 6

So find that du = 1/x dx And you should get \[\int\limits u du = 1/2u^2+c\] then you can just plug in u.

Answer 7

\[\large \int\limits \ln x \left(\color{#3366CF}{\frac{1}{x}dx}\right)\]

\[\large u=\ln x\]\[\large \color{#3366CF}{du=\frac{1}{x}dx}\]

Maybe the colors will help you see what's going on. I rewrote the integral so it's a little easier to identity your pieces.

Answer 8

Yah check out twists integral, it will end up being setup like that. :)

Answer 9

ok... so (1/2)(ln(x))^2+C

Answer 10

Yes

Answer 11

cool :) i didnt know u can draw here too

Answer 12

I understand now... du "should" replace the the other values and dx so you have a more simple integration on hand...

Answer 13

Exactly :D

Answer 14

let me see if i can do the next correctly and i'll have you check it if that's ok

Answer 15

ok so we have \[\int\limits_{}^{}x(4+x^2)^{10}dx\] = \[\int\limits_{}^{}(u)^{10}du = \frac{ (4+x^2) }{ 11 }+C\]

Answer 16

Your du should have a factor of 1/2 attached to it.

See how after you chose a \(\large u\), the derivative of that u is \(\large du=2x\;dx\) right?

Answer 17

u=4+x^2, du=2x dx, that's right

Answer 18

so where does the 2 go?

Answer 19

\[\large \int\limits\limits(4+x^2)^{10}\color{orangered}{x\;dx}\]\[\large u=4+x^2\]\[\large du=2\color{orangered}{x\;dx}\]Divide both sides by 2, to solve for x dx.

Answer 20

ok so \[2\int\limits_{}^{}(u)^{10}du\]