What is the integral of (2x-4)^{1/2}(x-2) ?

The integral of (2x-4)^{1/2}(x-2) is 2^{1/2}*2/5 (x-2)^{5/2}+C

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integral of (2x-4)^{1/2}(x-2)

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

2^{\frac{1}{2}} \cdot \frac{2}{5} (x - 2)^{\frac{5}{2}} + C

Solution steps

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

(2 x - 4)^{\frac{1}{2}} (x - 2) = 2^{\frac{1}{2}} (x - 2)^{\frac{1}{2}} (x - 2)

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

Take the constant out:

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

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

Simplify

(x - 2)^{\frac{1}{2}} (x - 2) : (x - 2)^{\frac{3}{2}}

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

Apply u-substitution

: \int u^{\frac{3}{2}} d u

= 2^{\frac{1}{2}} \cdot \int u^{\frac{3}{2}} d u

Apply the Power Rule

: \frac{2}{5} u^{\frac{5}{2}}

= 2^{\frac{1}{2}} \cdot \frac{2}{5} u^{\frac{5}{2}}

Substitute back

u = x - 2

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

Add a constant to the solution

= 2^{\frac{1}{2}} \cdot \frac{2}{5} (x - 2)^{\frac{5}{2}} + C

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What is the integral of (2x-4)^{1/2}(x-2) ?
The integral of (2x-4)^{1/2}(x-2) is 2^{1/2}*2/5 (x-2)^{5/2}+C