Wie binde ich $(\frac{1}{x^{4}}) d x$ $(1 / x ^ 4) dx$ ein?

$\int \frac{1}{x^{4}} . d x = - \frac{1}{3 x^{3}} + C$ $int 1/x^4color(white)(.)dx = -1/(3x^3) + C$

Beachten Sie, dass:

$\frac{d}{d x} \frac{1}{x^{3}} = \frac{d}{d x} x^{- 3} = - 3 x^{- 4} = - 3 (\frac{1}{x^{4}})$ $d/(dx) 1/x^3 = d/(dx) x^(-3) = -3 x^(-4) = -3(1/x^4)$

Damit:

$\int \frac{1}{x^{4}} . d x = - \frac{1}{3 x^{3}} + C$ $int 1/x^4color(white)(.)dx = -1/(3x^3) + C$

Wie binde ich (1 / x ^ 4) dx (1x4)dx (1 / x ^ 4) dx ein?