What is the integral from 0 to 3 of 9/(x-3) ?

The integral from 0 to 3 of 9/(x-3) is diverges

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integral from 0 to 3 of 9/(x-3)

\int_{0}^{3} \frac{9}{x - 3} d x

diverges

Solution steps

\int_{0}^{3} \frac{9}{x - 3} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 9 \cdot \int_{0}^{3} \frac{1}{x - 3} d x

Apply u-substitution

= 9 \cdot \int_{- 3}^{0} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 9 [ln ∣ u ∣]_{- 3}^{0}

Compute the boundaries:

diverges

= diverges

What is the integral from 0 to 3 of 9/(x-3) ?
The integral from 0 to 3 of 9/(x-3) is diverges