What is the integral of 4x^{-1/3} ?

The integral of 4x^{-1/3} is 6x^{2/3}+C

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integral of 4x^{-1/3}

\int 4 x^{- \frac{1}{3}} d x

6 x^{\frac{2}{3}} + C

Solution steps

\int 4 x^{- \frac{1}{3}} d x

Take the constant out:

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

= 4 \cdot \int x^{- \frac{1}{3}} d x

Apply the Power Rule

: \frac{3}{2} x^{\frac{2}{3}}

= 4 \cdot \frac{3}{2} x^{\frac{2}{3}}

Simplify

4 \cdot \frac{3}{2} x^{\frac{2}{3}} : 6 x^{\frac{2}{3}}

= 6 x^{\frac{2}{3}}

Add a constant to the solution

= 6 x^{\frac{2}{3}} + C

(2, 5), (1, 5)

\int_{2}^{5} 3 x d x

\int \frac{2 x - 5}{x ^{2} + 8} d x

\frac{d}{d x} (sinh (x^{2}))

f (x) = \frac{x ^{2} + x + 1}{x ^{2} + 1}