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

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

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

\frac{3}{8} x^{\frac{8}{3}} - 3 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 d x = 3 x

= \frac{3}{8} x^{\frac{8}{3}} - 3 x

Add a constant to the solution

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

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

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

\int sec^{2} (x) (1 + tan (x)) d x

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

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

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