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

The integral of ((2sqrt(x)-5)^2)/3 is 1/3 (2x^2-40/3 x^{3/2}+25x)+C

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

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

\frac{1}{3} (2 x^{2} - \frac{40}{3} x^{\frac{3}{2}} + 25 x) + C

Solution steps

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{3} \cdot \int (2 x - 5)^{2} d x

2 x - 5 = 5 (\frac{2 x}{5} - 1)

= \frac{1}{3} \cdot \int (5 (\frac{2 x}{5} - 1))^{2} d x

Simplify

(5 (\frac{2 x}{5} - 1))^{2} : 4 x - 20 x + 25

= \frac{1}{3} \cdot \int 4 x - 20 x + 25 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

= \frac{1}{3} (\int 4 x d x - \int 20 x d x + \int 25 d x)

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

\int 20 x d x = \frac{40}{3} x^{\frac{3}{2}}

\int 25 d x = 25 x

= \frac{1}{3} (2 x^{2} - \frac{40}{3} x^{\frac{3}{2}} + 25 x)

Add a constant to the solution

= \frac{1}{3} (2 x^{2} - \frac{40}{3} x^{\frac{3}{2}} + 25 x) + C

\int_{0}^{\frac{π}{6}} 3 cos^{5} (3 x) d x

\frac{d}{d y} (x^{2} + x y + y^{2})

\int_{0}^{4} 2 x - x d x

x \to 0 lim (\frac{1}{x ^{3} - 1} + 1)

x \to 0 lim (\frac{x ^{2} - 5}{x})

What is the integral of ((2sqrt(x)-5)^2)/3 ?
The integral of ((2sqrt(x)-5)^2)/3 is 1/3 (2x^2-40/3 x^{3/2}+25x)+C