#How to compute antiderivative in pic?
51 messages · Page 1 of 1 (latest)
I was thinking of doing the u' times u formula
but then arctg would have -1 as exponent which makes that formula not applicable
So what do?
<@&727457725017096242>
What is "arctg"? Do you mean "arctan"?
yea
arctg is also valid
same thing with sin and sen
It's not a notation I've ever seen.
EUbros represent
I dont think we're supposed to do u-sub for these exercises, I think these are supposed to have a simpler solution as they're more introductory also I haven't done much u-sub before so I'm not super confident with it
oh sure
Then isn't it also possible that this is supposed to be a problem where you practice u sub?
oh yh
I just saw what u have to do
@glacial axle try computing the derivative of arctg(x/2)
see what u get
you'll see a familiar expression
Holy shit, I'm an idiot. I already did that, but I did it wrong.
But if I separate it into two fractions (one multiplied by the other) and make the first fraction be the derivative of the second I still have the problem that the exponent of the second fraction would be -1 which makes the rule invalid.
You're trying to get to this form, right?
wdym?
nop
there's another form
lemme write it down
$\int { \frac{f^{'}(x)}{f(x)} dx} = ln(f(x)) + C$
Arm
it's really the absolute value, but u get the idea
I'm sorry, but I don't see how to get to that form... 😦
what's the derivative of ln(f(x))?
f'(x)/f(x)
f(x)/int(f(x))
?
Not sure what you mean both sides
u just get the integral of f'(x)/f(x) = ln(f(x) + c
ok and...?
so that's the form for n=-1
but still
u can do this without using it
but yh, this is the easiest way
and probably the intended way hahah
Let
U = arctan(x/2)*1/2
Du = 1/4 + x^2
$2\int \frac{du}{u} = 2\ln{|u|} + C= 2\ln{|\frac{1}{2}\arctan(\frac{x}{2})|} + C$