Solutions that give you a half chub

Problems when you first figured out how to solve them, you got hard

While this is standard, when you really start being able to on-the-fly recognize easy way to add zero or multiply by 1 it's an amazing tool that never stops paying dividends.

Keep it up OP.

What's your variable of integration?

Disgusting.

To show [math] e^{\pi i}=-1 [/math], note:

[math] e^{\pi i}*0=-1*0 [/math]

[math] 0=0 [/math]

Proved

Honestly my first inclination for an integral like this would to just do a substitution (u = x+1).

But I really enjoy the +1 - 1 tricks. A better example of this OP would be integral of x^2 over (x^2 +1)

how do you cancel the x with substitution?

If u = x+1, then du = dx, x = u -1 ( the numerator), and the denominator is u.

Then the integral is (u-1)/u, which is easy

Too lazy to do it right now, but this isn't as hard as it looks to prove.

literally nobody thinks it's hard to prove.

Honestly deriving boltzmanns gas equation for a challenge question in my calc 2 class got me going

[math]\tfrac{dy}{dx}\sqrt{x}= \tfrac{1}{2\sqrt{x}}[/math]

this is incorrect, but i might also be falling for b8

what is it then

>dy/dx
thats what's wrong, its d/dx

this user (who is not me) is correct

oh ya woops

Can you state the question? I want to give it a try.

qst : prove that every non constant complex polynomial is a surjection.

answer :

[math]P(z) = w[/math] is equivalent to [math](P-w)(z) = 0[/math]

d'Alembert-Gauss

done

Always loved this simple thing. Never understood why.

[math]\int \log (x^2+1)dx[/math] can be solved by partial integration. Define [math]dv = dx[/math] and [math]u = \log (x^2+1)[/math].
Then [math]\int \log (x^2+1)dx = x \log (x^2+1) - \int \frac{2x^2}{x^2 +1}dx[/math], which can readily be solved by some more partial integrations

it's a differential equation
y(x) = log(x)/2 + c

You could have just did polynomial division.

1*1
I always thought it was 2

Faddeev and Popov agree.

the Eckmann-Hilton argument

a^n=b^n+c^n where n>2

What the hell are you talking about.

ur mom

t-that's [math]y' + \sqrt{x} = \frac{1}{2\sqrt{x}}[/math]

You mean d/dx.

Nope, it's not.

If you know what cos() and sin() are it is completely obvious.