What is the integral from 3 to 5 of 1/(x^2-6x+9) ?

The integral from 3 to 5 of 1/(x^2-6x+9) is diverges

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integral from 3 to 5 of 1/(x^2-6x+9)

\int_{3}^{5} \frac{1}{x ^{2} - 6 x + 9} d x

diverges

Solution steps

\int_{3}^{5} \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_{3}^{5} \frac{1}{( x - 3 ) ^{2}} d x

Apply u-substitution

: \int_{0}^{2} \frac{1}{u ^{2}} d u

= \int_{0}^{2} \frac{1}{u ^{2}} d u

Apply the Power Rule

: [- \frac{1}{u}]_{0}^{2}

= [- \frac{1}{u}]_{0}^{2}

Compute the boundaries:

diverges

= diverges

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What is the integral from 3 to 5 of 1/(x^2-6x+9) ?
The integral from 3 to 5 of 1/(x^2-6x+9) is diverges