...
95% of you will fail
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damn, tough one.
guess it's something with arcsin (because of the form), but I don't have the patience to start substituting integrals.
Nice try. We won't do your homework, cunt.
Homework board is this way friendo
The amount of effort you put into this homework post is impressive, I'll say that.
>brainlets spotted who are omitting a cry of help because they started the problem and realized it's too hard for them so they project their insecurities onto the alpha who put them in this conundrum
wew lad
The indefinite integral of f(x) is:
F(x)=ln(tan(arcsin(x)-1))-(5/8)ln(tan^2(arcsin(x))+1)-(1/2)arcsin(x)+c
Next time give me something less easy ;)
Cleo pls
>omitting a cry of help
>omitting
Looks like a brainlet is you
You can't answer it, you never specified what to compute the integral in respect to.
>indefinite integral
That isn't a thing. Made up notion taught in brainlet tier calc courses.
Are you claiming antiderivatives aren't real math? Care to back up that claim?
When you differentiate your result you don't get the original function. How does it feel to be this wrong?
There is no inverse to the derivative operator. Finding antiderivatives is just an adhoc system of calculation tools and is in general an ill-defined notion.
And "indefinite integral" is just a flat out horrible abuse of terminology.
1 over x fuckface
>no inverse to the derivative operator
???
(log(-1 + sqrt(5) - 2 x) - log(1 + sqrt(5) + 2 x))/sqrt(5)
now fuck off cunt
1/4 (-2 sin^(-1)(x) - 2 tanh^(-1)(x/sqrt(1 - x^2)) + ln(1 - 2 x^2))
You sure didn't pay attention to calc.
tutorial.math.lamar.edu
study hard brainlet
lmaooooo
There is no inverse to the derivative operator.
Calc classes are horrible excuses for mathematics
That "definition" is not at all rigorous
>There is no inverse to the derivative operator.
There is if you consider functions modulo an added constant. Don't be such a butthurt faggot