What is the antiderivative of $\frac{3}{x^{2}}$ $3/x^2$ ?

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What is the antiderivative of $\frac{3}{x^{2}}$ $3/x^2$ ?

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Answer 1 · 2016-10-12T22:14:04

$\int \frac{3}{x^{2}} d x = - \frac{3}{x} + C$ $int 3/x^2 dx = -3/x + C$

Explanation:

$\frac{3}{x^{2}} = 3 x^{- 2}$ $3/x^2 = 3x^(-2)$

The power rule of differentiation tells us that:

$\frac{d}{d x} x^{n} = n x^{n - 1}$ $d/(dx) x^n = n x^(n-1)$

Hence we find:

$\frac{d}{d x} \frac{1}{x} = \frac{d}{d x} x^{- 1} = (- 1) x^{- 2} = - \frac{1}{x^{2}}$ $d/(dx) 1/x = d/(dx) x^(-1) = (-1) x^(-2) = -1/x^2$

Hence:

$\frac{d}{d x} (- \frac{3}{x}) = (- 3) \frac{d}{d x} (\frac{1}{x}) = (- 3) (- \frac{1}{x^{2}}) = \frac{3}{x^{2}}$ $d/dx (-3/x) = (-3) d/(dx) (1/x) = (-3) (-1/x^2) = 3/x^2$

So:

$\int \frac{3}{x^{2}} d x = - \frac{3}{x} + C$ $int 3/x^2 dx = -3/x + C$

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What is the antiderivative of $\frac{3}{x^{2}}$ $3/x^2$ ?

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