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

The integral of x(5x-3)^2 is (25x^4)/4-10x^3+(9x^2)/2+C

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integral of x(5x-3)^2

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

\frac{25 x ^{4}}{4} - 10 x^{3} + \frac{9 x ^{2}}{2} + C

Solution steps

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

Factor

5 x - 3 : - (- 5 x + 3)

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

(- (- 5 x + 3))^{2} = (- 5 x + 3)^{2}

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

Expand

x (- 5 x + 3)^{2} : 25 x^{3} - 30 x^{2} + 9 x

= \int 25 x^{3} - 30 x^{2} + 9 x 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 25 x^{3} d x - \int 30 x^{2} d x + \int 9 x d x

\int 25 x^{3} d x = \frac{25 x ^{4}}{4}

\int 30 x^{2} d x = 10 x^{3}

\int 9 x d x = \frac{9 x ^{2}}{2}

= \frac{25 x ^{4}}{4} - 10 x^{3} + \frac{9 x ^{2}}{2}

Add a constant to the solution

= \frac{25 x ^{4}}{4} - 10 x^{3} + \frac{9 x ^{2}}{2} + C

\int e^{3 x} + \frac{1}{x} d x

\frac{d}{d x} (x \frac{d y}{d x})

y^{^{'}} + 3 y = 2 e^{- 3 t}

f (x) = \frac{( x - 2 )}{( 2 x ^{2} + 3 )}

\int_{2}^{3} \frac{1}{x ^{2} - 4 x} d x

What is the integral of x(5x-3)^2 ?
The integral of x(5x-3)^2 is (25x^4)/4-10x^3+(9x^2)/2+C