What is the integral of (x^{1/3}-2x^{-2/3}+3) ?

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

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integral of (x^{1/3}-2x^{-2/3}+3)

\int (x^{\frac{1}{3}} - 2 x^{- \frac{2}{3}} + 3) d x

\frac{3}{4} x^{\frac{4}{3}} - 6 x^{\frac{1}{3}} + 3 x + C

Solution steps

\int x^{\frac{1}{3}} - 2 x^{- \frac{2}{3}} + 3 d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int x^{\frac{1}{3}} d x - \int 2 x^{- \frac{2}{3}} d x + \int 3 d x

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

\int 2 x^{- \frac{2}{3}} d x = 6 x^{\frac{1}{3}}

\int 3 d x = 3 x

= \frac{3}{4} x^{\frac{4}{3}} - 6 x^{\frac{1}{3}} + 3 x

Add a constant to the solution

= \frac{3}{4} x^{\frac{4}{3}} - 6 x^{\frac{1}{3}} + 3 x + C

\int (\frac{1}{x} + \frac{2}{x ^{2}} + \frac{3}{x ^{3}}) d x

\int \frac{7}{x} d x

y (1 + x y) d x + (2 y - x) d y = 0

\int (\frac{2}{x}) d x

f (x) = 6 x^{3} - 6 x^{2} + 4

What is the integral of (x^{1/3}-2x^{-2/3}+3) ?
The integral of (x^{1/3}-2x^{-2/3}+3) is 3/4 x^{4/3}-6x^{1/3}+3x+C