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

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

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

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

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

Solution steps

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

Expand

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

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

\int 16 d x = 16 x

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

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

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

Add a constant to the solution

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

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