What is the integral of (3x^5-2x^3) ?

The integral of (3x^5-2x^3) is (x^6)/2-(x^4)/2+C

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integral of (3x^5-2x^3)

\int (3 x^{5} - 2 x^{3}) d x

\frac{x ^{6}}{2} - \frac{x ^{4}}{2} + C

Solution steps

\int 3 x^{5} - 2 x^{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 3 x^{5} d x - \int 2 x^{3} d x

\int 3 x^{5} d x = \frac{x ^{6}}{2}

\int 2 x^{3} d x = \frac{x ^{4}}{2}

= \frac{x ^{6}}{2} - \frac{x ^{4}}{2}

Add a constant to the solution

= \frac{x ^{6}}{2} - \frac{x ^{4}}{2} + C

cos (tan (x))

\frac{\partial}{\partial x} (e^{x} - x y^{3})

y = - 2 x^{2}

\frac{x ^{2}}{3 + 4 x}

t \cdot e^{- 3 t} \cdot sin (5 t)

What is the integral of (3x^5-2x^3) ?
The integral of (3x^5-2x^3) is (x^6)/2-(x^4)/2+C