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

The integral of (sqrt(x)-4)^2 is (x^2)/2-16/3 x^{3/2}+16x+C

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integral of (sqrt(x)-4)^2

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

\frac{x ^{2}}{2} - \frac{16}{3} x^{\frac{3}{2}} + 16 x + C

Solution steps

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

Expand

(x - 4)^{2} : x - 8 x + 16

= \int x - 8 x + 16 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 d x - \int 8 x d x + \int 16 d x

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

\int 8 x d x = \frac{16}{3} x^{\frac{3}{2}}

\int 16 d x = 16 x

= \frac{x ^{2}}{2} - \frac{16}{3} x^{\frac{3}{2}} + 16 x

Add a constant to the solution

= \frac{x ^{2}}{2} - \frac{16}{3} x^{\frac{3}{2}} + 16 x + C

y^{^{''}} + 4 y^{^{'}} = 17 cos (t), y (0) = - 1, y^{^{'}} (0) = 0

\frac{d}{d x} (- \frac{16}{( 1 + 4 x ) ^{2}})

\int_{1}^{2} 21 t^{2} d t

\frac{d}{d x} (x - 8)

4^{\frac{x}{l n ( 2 )}}, at x = ln (16)

What is the integral of (sqrt(x)-4)^2 ?
The integral of (sqrt(x)-4)^2 is (x^2)/2-16/3 x^{3/2}+16x+C