/sci/ - Science & Math

Why is what though?

Why (1/x)dx = ln x

http://www.youtube.com/watch?v=yUpDRpkUhf4

Found it!

DERPIN' THROUGH CALCULUS 1 FOR GREAT JUSTICE!

bump. why is that shit though!

<span class="math">y=ln\ x[/spoiler]
So <span class="math">x=e^{y}[/spoiler]
Therefore <span class="math">\frac{dx}{dy} = e^{y}[/spoiler]
So <span class="math">\frac{dy}{dx} = \frac{1}{e^{y}}[/spoiler]
But <span class="math">x = e^{y}[/spoiler] so <span class="math"> \frac{dy}{dx} = \frac{1}{x}[/spoiler]

Make sense?

The integral of <span class="math">x^n[/spoiler] should be <span class="math">\frac{x^{n+1}}{n}+c[/spoiler]. For <span class="math">n=-1[/spoiler], that clearly doesn't work, since then the "antiderivative" is not x-dependend anymore, which is certainly wrong.

Now what is the integral of <span class="math">\frac{1}{x}[/spoiler]. It's problematic to argue using an integral from <span class="math">0[/spoiler] to <span class="math">a[/spoiler], because for <span class="math">x=0[/spoiler], the integral will diverge.

To compute the integral, shift the whole function to the left by a constant, i.e. consider <span class="math">\frac{1}{x-1}[/spoiler].
This is minus the function which is expanded as a the geometric series
http://en.wikipedia.org/wiki/Geometric_series#Formula

check out this
http://www.wolframalpha.com/input/?i=Series[-1%2F%28x-1%29%2C{x%2C0%2C6}]

the integral is that function
http://www.wolframalpha.com/input/?i=Series[Log[x-1]%2C{x%2C0%2C6}]

Okay another problem:
xdx/(4+x^4)
Rules: Must not use integration by parts:
(4+x^4)=t
x=(t-4)^(1/4)
dx=dt/4(t-4)^(3/4)
...
...
...
(sqrt(t-4))dt/4t, assuming 1/4 is out of the integral=> sqrt(t-4)dt/t.

Now what?

Okay, trying another substitution.
1/4 integral sqrt(t-4)dt/t
(sqrt(t-4))=k
t=k^2+4
dt=2k
So that leads me to...
1/4 integral k*2k/(k^2+4)
pulling 2 out of the integral

1/2 integral k^2/(k^2+4)

Seems a lot more simple...but still can't see the answer.

The derivative of tan^-1 x is 1/(x^2+1), but you have x^4+4, luckily theres an x up there that shows up applying chain rule so the integral is

(1/2)tan^-1(x^2/2)

No method needed, if you cant see it try the substitution x^2=t

Yay, got it with x^2=t
Thanks guys...Just about 30 more to go...and the night is long...Exam is in 2 weeks! Woo!

So, Cya later on our nighly mathlover's programme, derpchan!

>>4280483
*nightly

No, I'm not drunk.
I wish I was :(