Antiderivative?

Question

What is the antiderivative of:

1) x √x

2) 1/ x^2

Thanks, I am a little stuck here.

Answer 1

Remember the general formula

Anti derivative of x^a = x^(a+1)/(a+1) +C, a can be any real number but -1

So instead of x√x, write x^(3/2), its antiderivative is x^(5/2) / (5/2) +C = (2/5)x^2.√x + C

Instead of 1/x^2, write x^(-2), its antiderivative is x^(-1)/(-1) +C  = -1/x + C

Answer 2

∫ x √x dx = ∫ x ^(3/2)dx = (2/5)x^(3/2+1)=(2/5)x^(5/2) +C

2)∫ 1/x^2 dx =∫ x^-2 dx = x^-2+1 /(-2+1) = -x^-1 = -1/x+C

Answer 3

antiderivative = intergration

hope u know that that is lol

Answer 4

This is just an application of the power law for antidifferentiation, which says that Integral[x^n dx] = [x^(n + 1)]/[n + 1] + C.

1) x√x = x^(3/2), => Integral[x^(3/2) dx] = [x^(5/2)]/[5/2] = (2/5)x^2√x + C.

2) 1/(x^2) = x^(-2), => Integral[x^(-2) dx] = [x^(-1)]/[-1] = -1/x + C.