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

The integral of (sqrt(x-4))^2 is (x^2)/2-4x+C

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

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

\frac{x ^{2}}{2} - 4 x + C

Solution steps

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

Simplify

(x - 4)^{2} : x - 4

= \int x - 4 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 4 d x

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

\int 4 d x = 4 x

= \frac{x ^{2}}{2} - 4 x

Add a constant to the solution

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

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