What is the integral of 1/(x^2-6x+9) ?

The integral of 1/(x^2-6x+9) is -1/(x-3)+C

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integral of 1/(x^2-6x+9)

\int \frac{1}{x ^{2} - 6 x + 9} d x

- \frac{1}{x - 3} + C

Solution steps

\int \frac{1}{x ^{2} - 6 x + 9} d x

Take the partial fraction of

\frac{1}{x ^{2} - 6 x + 9} : \frac{1}{( x - 3 ) ^{2}}

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

Apply u-substitution

: \int \frac{1}{u ^{2}} d u

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

Apply the Power Rule

: - \frac{1}{u}

= - \frac{1}{u}

Substitute back

u = x - 3

= - \frac{1}{x - 3}

Add a constant to the solution

= - \frac{1}{x - 3} + C

\int \frac{1}{2} x^{- \frac{1}{2}} d x

\int x e^{- 0.1 x} d x

\int (2 x^{4} + 5 x^{3} - x + 5) d x

\int (12 x^{2} - 6 x + 4) d x

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

What is the integral of 1/(x^2-6x+9) ?
The integral of 1/(x^2-6x+9) is -1/(x-3)+C