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

The integral of (2x-sqrt(x))^2 is (4x^3)/3-8/5 x^{5/2}+(x^2)/2+C

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

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

\frac{4 x ^{3}}{3} - \frac{8}{5} x^{\frac{5}{2}} + \frac{x ^{2}}{2} + C

Solution steps

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

Expand

(2 x - x)^{2} : 4 x^{2} - 4 x^{\frac{3}{2}} + x

= \int 4 x^{2} - 4 x^{\frac{3}{2}} + 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 4 x^{2} d x - \int 4 x^{\frac{3}{2}} d x + \int x d x

\int 4 x^{2} d x = \frac{4 x ^{3}}{3}

\int 4 x^{\frac{3}{2}} d x = \frac{8}{5} x^{\frac{5}{2}}

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

= \frac{4 x ^{3}}{3} - \frac{8}{5} x^{\frac{5}{2}} + \frac{x ^{2}}{2}

Add a constant to the solution

= \frac{4 x ^{3}}{3} - \frac{8}{5} x^{\frac{5}{2}} + \frac{x ^{2}}{2} + C

\frac{\partial}{\partial x} (4 sin (x) cos (y))

\frac{\partial}{\partial K} (L^{\frac{1}{2}} K^{\frac{1}{2}})

x \to 4 lim (x + 4)

f (x) = 4 x - 7

\int_{0}^{4} y d y

What is the integral of (2x-sqrt(x))^2 ?
The integral of (2x-sqrt(x))^2 is (4x^3)/3-8/5 x^{5/2}+(x^2)/2+C