What is the integral of (8/(x^3)) ?

The integral of (8/(x^3)) is -4/(x^2)+C

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integral of (8/(x^3))

\int (\frac{8}{x ^{3}}) d x

- \frac{4}{x ^{2}} + C

Solution steps

\int \frac{8}{x ^{3}} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 8 \cdot \int \frac{1}{x ^{3}} d x

Apply the Power Rule

: - \frac{1}{2 x ^{2}}

= 8 (- \frac{1}{2 x ^{2}})

Simplify

8 (- \frac{1}{2 x ^{2}}) : - \frac{4}{x ^{2}}

= - \frac{4}{x ^{2}}

Add a constant to the solution

= - \frac{4}{x ^{2}} + C

f (x) = 2 x^{3} - 3 x^{2} - 16 x + 3, at x = - 4

\frac{d}{d x} ((e^{x})^{s i n (x)})

y = x^{3} - 7 x^{2} + 16 x + 7

(0, - 2), (5, - 4)

\int_{0}^{3} 2 e^{- 3 x} d x