How do I find these ANTI-DERIVATIVES?

We get …

3 (1/3) x^3 + c = x^3 + c

2. m(x) = (-2/y^3)

We re-write it as …

-2 y^-3

and get …

-2 (-1/2) y^-2 + c

which becomes …

y^-2 + c

which becomes …

1 / y^2 + c

3. g(x) = x ^ π

Treat π as a constant (such as ‘2’) and we get …

1/(π+1) x^(π+1) + c

4. h(x) = cos x

The derivative of Sin(x) is Cos(x) hence we get …

sin(x) + c

5. p(x) = 1/ (x^2 + 1)

The derivative of arctan(x) is 1/(x^2 + 1) hence we get …

arctan(x) + c

Arctan(x) means 1/tan(x)

1. f(x) = 3x^2
Is the anti-derivative x^3 + c?
yes
2. m(x) = (-2/y^3)
Is the anti-derivative (1/2) y^-4 + c?
you have a confusion of x’s and y’s…

-2y^-3 would become y^-2

I’m not really sure anymore how to find the rest of these anti-derivatives.
3. g(x) = x ^ π

Just like x^any number… don’t let it throw you that it is not an integer.

G(x) = (1/(π+1))x ^ (π+1) +c

4. h(x) = cos x
H(x) = sin x +c .. I don’t have the space here to derive this one… the derivative of sin x = cos x the derivative of cos x = -sin x

5. p(x) = 1/ (x^2 + 1)

P(x) = tan^-1 x +c

And this one is even more difficult to explain here…

you could get tricky with it, too…

1/ (x^2 + 1) = 1/(x+i)(x-i) = 1/2i (1/(x+i) – 1/(x-i))
P = 1/2 i (ln(x+i) – ln(x-i))+c = 1/2 i ln(x+i)/(x-i) +c

but inverse tangent is the safer way to go…

1. correct
2. Why that is brilliant! -4 = -3 + 1 ….I wish I had your skills in addition
3. 3.1415926… + 1 = 4.1415926… what is your problem?
4. oh, I see… either you don’t know the differential for sin(x) or you have ADD or you are just lazy.
5. (x²+1)⁻¹ … apparently you don’t know the differentials for the inverse trig functions either…

3) = x^ ( pi + 1)
…… —————- + c …………………………….. Power rule
……… pi + 1

4) = -sin x + c

5) = arc tan x + c