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

The integral of (x^{3/2}-5x)^2 is (x^4)/4-20/7 x^{7/2}+(25x^3)/3+C

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integral of (x^{3/2}-5x)^2

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

\frac{x ^{4}}{4} - \frac{20}{7} x^{\frac{7}{2}} + \frac{25 x ^{3}}{3} + C

Solution steps

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

Expand

(x^{\frac{3}{2}} - 5 x)^{2} : x^{3} - 10 x^{\frac{5}{2}} + 25 x^{2}

= \int x^{3} - 10 x^{\frac{5}{2}} + 25 x^{2} 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 x^{3} d x - \int 10 x^{\frac{5}{2}} d x + \int 25 x^{2} d x

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

\int 10 x^{\frac{5}{2}} d x = \frac{20}{7} x^{\frac{7}{2}}

\int 25 x^{2} d x = \frac{25 x ^{3}}{3}

= \frac{x ^{4}}{4} - \frac{20}{7} x^{\frac{7}{2}} + \frac{25 x ^{3}}{3}

Add a constant to the solution

= \frac{x ^{4}}{4} - \frac{20}{7} x^{\frac{7}{2}} + \frac{25 x ^{3}}{3} + C

\frac{d}{d x} (sin^{3} (2 x - 3))

\frac{d y}{d t} + e^{t} y = 13 e^{t}

\frac{1}{4 + x ^{2}}, 0

\frac{d}{d x} (x^{3} - 4 x + 6)

\int x arctan (5 x) d x

What is the integral of (x^{3/2}-5x)^2 ?
The integral of (x^{3/2}-5x)^2 is (x^4)/4-20/7 x^{7/2}+(25x^3)/3+C