What is the integral of x^2(2x^3-5)^4 ?

The integral of x^2(2x^3-5)^4 is (16x^{15}}{15}-\frac{40x^{12})/3+(200x^9)/3-(500x^6)/3+(625x^3)/3+C

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integral of x^2(2x^3-5)^4

\int x^{2} (2 x^{3} - 5)^{4} d x

\frac{16 x ^{15}}{15} - \frac{40 x ^{12}}{3} + \frac{200 x ^{9}}{3} - \frac{500 x ^{6}}{3} + \frac{625 x ^{3}}{3} + C

Solution steps

\int x^{2} (2 x^{3} - 5)^{4} d x

Expand

x^{2} (2 x^{3} - 5)^{4} : 16 x^{14} - 160 x^{11} + 600 x^{8} - 1000 x^{5} + 625 x^{2}

= \int 16 x^{14} - 160 x^{11} + 600 x^{8} - 1000 x^{5} + 625 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 16 x^{14} d x - \int 160 x^{11} d x + \int 600 x^{8} d x - \int 1000 x^{5} d x + \int 625 x^{2} d x

\int 16 x^{14} d x = \frac{16 x ^{15}}{15}

\int 160 x^{11} d x = \frac{40 x ^{12}}{3}

\int 600 x^{8} d x = \frac{200 x ^{9}}{3}

\int 1000 x^{5} d x = \frac{500 x ^{6}}{3}

\int 625 x^{2} d x = \frac{625 x ^{3}}{3}

= \frac{16 x ^{15}}{15} - \frac{40 x ^{12}}{3} + \frac{200 x ^{9}}{3} - \frac{500 x ^{6}}{3} + \frac{625 x ^{3}}{3}

Add a constant to the solution

= \frac{16 x ^{15}}{15} - \frac{40 x ^{12}}{3} + \frac{200 x ^{9}}{3} - \frac{500 x ^{6}}{3} + \frac{625 x ^{3}}{3} + C

y^{^{'}} = \frac{5}{y}

\frac{d ^{3}}{d x ^{3}} (2 t + 8)

t sin (2 t)

\frac{d}{d h} (\frac{n ^{2} h ^{2}}{8 m L ^{2}})

t^{2} y^{^{''}} + 19 t y^{^{'}} + 145 y = 0

What is the integral of x^2(2x^3-5)^4 ?
The integral of x^2(2x^3-5)^4 is (16x^{15}}{15}-\frac{40x^{12})/3+(200x^9)/3-(500x^6)/3+(625x^3)/3+C