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

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

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integral of (x^2-4x+4)^{3/4}

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

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

Solution steps

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

Complete the square

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

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

Apply u-substitution

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

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

Apply the Power Rule

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

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

Substitute back

u = x - 2

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

Add a constant to the solution

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

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