What is the integral of \sqrt[-3]{x}+2x ?

The integral of \sqrt[-3]{x}+2x is 3/2 x^{2/3}+x^2+C

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integral of \sqrt[-3]{x}+2x

\frac{3}{2} x^{\frac{2}{3}} + x^{2} + 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 2 x d x = x^{2}

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

Add a constant to the solution

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

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

x \to 2 lim (F (x))

\frac{d}{d x} ((6 x^{2} + 7) (6 x + x))

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

y^{^{''}} + y = 3 sec^{3} (2 t)

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