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

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

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

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

\frac{x ^{6}}{3} - \frac{3 x ^{4}}{4} + C

Solution steps

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

Expand

x^{3} (2 x^{2} - 3) : 2 x^{5} - 3 x^{3}

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

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

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

= \frac{x ^{6}}{3} - \frac{3 x ^{4}}{4}

Add a constant to the solution

= \frac{x ^{6}}{3} - \frac{3 x ^{4}}{4} + C

\int (\frac{x}{3})^{3} d x

x \to 1 lim (\frac{7}{x ^{3} - 1})

4 t^{- \frac{3}{8}} + 5 ln (t) + ln (5)

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

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