What is the integral of (x^2-4x+4)^{10} ?

The integral of (x^2-4x+4)^{10} is ((x-2)^{21})/(21)+C

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integral of (x^2-4x+4)^{10}

\int (x^{2} - 4 x + 4)^{10} d x

\frac{( x - 2 ) ^{21}}{21} + C

Solution steps

\int (x^{2} - 4 x + 4)^{10} d x

Complete the square

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

= \int ((x - 2)^{2})^{10} d x

Apply u-substitution

: \int u^{20} d u

= \int u^{20} d u

Apply the Power Rule

: \frac{u ^{21}}{21}

= \frac{u ^{21}}{21}

Substitute back

u = x - 2

= \frac{( x - 2 ) ^{21}}{21}

Add a constant to the solution

= \frac{( x - 2 ) ^{21}}{21} + C

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What is the integral of (x^2-4x+4)^{10} ?
The integral of (x^2-4x+4)^{10} is ((x-2)^{21})/(21)+C