What is the integral of (3x^2-2/(x^3)+\sqrt[3]{x}-8) ?

The integral of (3x^2-2/(x^3)+\sqrt[3]{x}-8) is x^3+1/(x^2)+3/4 x^{4/3}-8x+C

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integral of (3x^2-2/(x^3)+\sqrt[3]{x}-8)

x^{3} + \frac{1}{x ^{2}} + \frac{3}{4} x^{\frac{4}{3}} - 8 x + C

Solution steps

Apply the Sum Rule:

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

\int 3 x^{2} d x = x^{3}

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

\int 8 d x = 8 x

= x^{3} - (- \frac{1}{x ^{2}}) + \frac{3}{4} x^{\frac{4}{3}} - 8 x

Simplify

= x^{3} + \frac{1}{x ^{2}} + \frac{3}{4} x^{\frac{4}{3}} - 8 x

Add a constant to the solution

= x^{3} + \frac{1}{x ^{2}} + \frac{3}{4} x^{\frac{4}{3}} - 8 x + C

\frac{2 x}{x ^{2} + 1}

x^{^{''}} + 3 x^{^{'}} + 2 x = e^{- t}

(3 x + 25)

\frac{\partial}{\partial x} (e^{2 x} cos (y))

x \to - 2 lim (3 x + 4)

What is the integral of (3x^2-2/(x^3)+\sqrt[3]{x}-8) ?
The integral of (3x^2-2/(x^3)+\sqrt[3]{x}-8) is x^3+1/(x^2)+3/4 x^{4/3}-8x+C