What is the integral from 0 to 2 of 7/(x-2) ?

The integral from 0 to 2 of 7/(x-2) is diverges

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integral from 0 to 2 of 7/(x-2)

\int_{0}^{2} \frac{7}{x - 2} d x

diverges

Solution steps

\int_{0}^{2} \frac{7}{x - 2} d x

Take the constant out:

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

= 7 \cdot \int_{0}^{2} \frac{1}{x - 2} d x

Apply u-substitution

= 7 \cdot \int_{- 2}^{0} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 7 [ln ∣ u ∣]_{- 2}^{0}

Compute the boundaries:

diverges

= diverges

k = 1 \sum \infty ln (k)

t an g e n t f (x) = arcsin (3 x), at x = \frac{1}{6}

\int 1 + (2 x - \frac{1}{8 x})^{2} d x

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

(e^{12 x})^{^{'}}

What is the integral from 0 to 2 of 7/(x-2) ?
The integral from 0 to 2 of 7/(x-2) is diverges