/sci/ - Science & Math

>>3074574
Fuck you for reminding me that I don't remember anything from my calculus course last summer!

>>3074598
Hahah. Sorry. Maybe. Depends on how much you paid to be in that calculus course...

OP, can you explain what integration method you used to get your result?

>>3074633
Partial fractions.

Does this help:
http://www.wolframalpha.com/input/?i=int%281%2F%28x%5E4%2B1%29%29&a=*C.int-_*IntegralsWord-

(1/4x^3)Ln(x^4+1) ...you jelly?

But really i have no clue where to start... Dunno how to factor the bottom..if i knew that i might be able to figure it out.

>>3074642
What was your partial fraction decomposition?

>>3074644
I've been staring at that one for a while. They say "rewrite the integrand as ..." and... well, that doesn't really help me know why.

>>3074651
Yeah. Same here bro.

>>3074651
if it was x^4 -1 i think you could factor, but not x^4 + 1

i didn't learn how to solve this shit in calc 1 i hope it never pops up unless i get to use wolfram

>>3074657
[(Ax+b)/((x² - (√2)x + 1)] + [Cx+d((x² - (√2)x + 1)]=1

So
A=-(√2)/4
B=1/2
C=(√2)/4
D=1/2

Need more?

Hey homosex: use a fucking inverse trig function.

<span class="math">\displaystyle \int \frac{1}{a^2 + u^2}du = \frac{1}{a}arctan\frac{u}{a} + C[/spoiler]

bro you are supposed to use natural logs for this

u = x4 + 1

integral of 1/u = ln(u)du

the integral of 1/(x^2+1) is atan(x)

you can use this with the substitution y = x^2

>>3074675
You have to complete the square so that you can factor. I'm in calc one. Horseshit instructor thinks this is great for us.

>>3074690
Mfw fuck me for forgetting about inverse tan

>>3074675

No that's good. I'm only done with calc 2 so I'm not well versed in many of these shortcut integration methods. Thanks tho

Let u = (X^4)+1 and du=dx so that you have the integral of du/u.

Integrated, you get ln(u)+C. Plug (x^4)+1 back in for u and you get ln((x^4)+1)+C

>>3074690
>>3074689

Lol.
No.
Nice try.
Go here: http://www.wolframalpha.com/input/?i=1+%2F%281%2B+x^4%29

>>3074689

du = 4x^3

>>3074705
There are multiple ways to solve the same thing. Plug in numbers and see that you get the same answer.

>>3074706

OP didn't learn U-sub yet.

To all the faggots saying arc tan.

The answer of this integral is:
(-√2)/8[ln|x² - (√2)x + 1| - ln|(x² + (√2)x + 1)| -2[arctan((√2)x +1)+ arctan((√2)x -1)]]

>>3074717
Which is by far the easiest and most important tool when integrating.

>>3074714

Okay. Then I feel less idiotic.

But regardless. I still have to prove it this way.

And my question isn't even about the answer, tis about the step (¼)∫((-√2)x + 2)/(x² - (√2)x + 1) split into ((-√2)/8) ∫(2x - (√2))/(x² - (√2)x + 1) + (¼)∫1/(x² - (√2)x + 1).

>>3074717
Herp derp.

>>3074717
>>3074730
>implying you can usub this one

Bamp.

Looks like a good time to use integration by parts.

>>3074702
>du=dx
lolwut? Not in this case bro. If <span class="math">u=x^4+1[/spoiler] then
<div class="math"> \frac{du}{dx}=4x^3 \Rightarrow du=4x^3 dx </div>

>>3074812

Exactly. And where the fuck in that equation do you have a <span class="math">\displaystyle 4x^3[/spoiler], OP?

Fucking faggots I swear...

>>3074834
>lrn2differentiation

>>3075177

This is integration.

>>3075177
LOOOOOOOL

Bump for lulz value.

>>3074702
implying the derivative of x^4 + 1 is dx and not 4x^3
this is clearly an arctan

>>3075290
No, THIS IS SPARTA!!!

>>3075745
No, this is Patrick!