What is the integral of 3/2 x^2-\sqrt[3]{x^4} ?

The integral of 3/2 x^2-\sqrt[3]{x^4} is (x^3)/2-3/7 x^{7/3}+C

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integral of 3/2 x^2-\sqrt[3]{x^4}

\frac{x ^{3}}{2} - \frac{3}{7} x^{\frac{7}{3}} + 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 \frac{3}{2} x^{2} d x = \frac{x ^{3}}{2}

= \frac{x ^{3}}{2} - \frac{3}{7} x^{\frac{7}{3}}

Add a constant to the solution

= \frac{x ^{3}}{2} - \frac{3}{7} x^{\frac{7}{3}} + C

- x^{2} + x + a

\frac{d}{d x} (8 x^{5} + 45 x^{4} + 80 x^{3} + 50 x^{2} + 200)

e^{t} cos (t)

(x - 6) y^{^{'}} + y = x^{2} - 20

\int x^{3} 16 - x^{2} d x

What is the integral of 3/2 x^2-\sqrt[3]{x^4} ?
The integral of 3/2 x^2-\sqrt[3]{x^4} is (x^3)/2-3/7 x^{7/3}+C