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

The integral from 1 to 2 of 1/(x^2-6x+9) is 1/2

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

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

\frac{1}{2}

Decimal Notation

0.5

Solution steps

\int_{1}^{2} \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_{1}^{2} \frac{1}{( x - 3 ) ^{2}} d x

Apply u-substitution

: \int_{- 2}^{- 1} \frac{1}{u ^{2}} d u

= \int_{- 2}^{- 1} \frac{1}{u ^{2}} d u

Apply the Power Rule

: [- \frac{1}{u}]_{- 2}^{- 1}

= [- \frac{1}{u}]_{- 2}^{- 1}

Compute the boundaries:

\frac{1}{2}

= \frac{1}{2}

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