What is the integral of sqrt(x^2-6x+9) ?

The integral of sqrt(x^2-6x+9) is ((x-3)^2)/2+C

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integral of sqrt(x^2-6x+9)

\int x^{2} - 6 x + 9 d x

\frac{( x - 3 ) ^{2}}{2} + C

Solution steps

\int x^{2} - 6 x + 9 d x

Complete the square

x^{2} - 6 x + 9 : (x - 3)^{2}

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

Apply u-substitution

: \int u d u

= \int u d u

Apply the Power Rule

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

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

Substitute back

u = x - 3

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

Add a constant to the solution

= \frac{( x - 3 ) ^{2}}{2} + C

\int x (x - 3)^{5} d x

y = 1 + 3 x, 3 \leq x \leq 9

y = 10 x^{2}, y = x^{2} + 12

\int e^{\frac{- 4 x}{3}} d x

x \to 0 lim (e^{x} - 1 - x)

What is the integral of sqrt(x^2-6x+9) ?
The integral of sqrt(x^2-6x+9) is ((x-3)^2)/2+C