Can 4chan integrate this?

>>8896784
No

>>8896784
Sub in tan for x^2 or some shit and then divide by the exponent on the triangles hypotenuse. Did I get this right?

>>8896784
No because this has a non-elementary integral.

>>8896784
Yes, get the Taylor series and integrate.

>>8896825
Let me know which point you choose for the Taylor series such that the radius of convergence precisely covers [math](0,\infty)[/math]

>>8896784
https://www.youtube.com/watch?v=Al6p3L2l9zY

>>8896784

http://www.integral-calculator.com/

[eqn]
4\, \pi \operatorname{arccot} \sqrt{\phi}
[/eqn]
t. Cleo

>>8896920
http://www.wolframalpha.com/input/?i=derivative+of+4+pi+arccot(phi%5E0.5)

What kind of chinese bootleg version of Cleo are you?

File: Capture.png (127KB, 1106x793px) Image search: [Google]

127KB, 1106x793px

>>8896784
No

File: chen_stuck.png (23KB, 388x258px)

23KB, 388x258px

>>8896784
[math]x \rightarrow x^2 \implies \int_0^\infty \frac{\cos(1-x^2)}{x^2 + \sqrt{x}} dx \rightarrow 2\int_0^\infty \frac{\cos(1-x^4)}{x^3 + 1}dx = \int_{-\infty}^{\infty} \frac{\cos(1-x^4)}{x^3 + 1}dx = \sum_{\operatorname{Res} k}\cos(1-x_k^4) =
\sum_{n=0}^{2} \operatorname{Re}\left[ \cos(1-e^{i\frac{4}{3}n\pi})\right][/math]

>>8898303
I love you.

>>8898303
wtf did you do there

>>8898303
Wrong. If anything, you should have divided by 2x, not multiplied.

>>8898303
>[eqn]2\int_0^\infty \frac{\cos(1-x^4)}{x^3 + 1}dx = \int_{-\infty}^{\infty} \frac{\cos(1-x^4)}{x^3 + 1}dx[/eqn]
But the integrand is not an even function you fucking retard