What is the integral from 2 to 3 of 1/(y-3) ?

The integral from 2 to 3 of 1/(y-3) is diverges

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integral from 2 to 3 of 1/(y-3)

\int_{2}^{3} \frac{1}{y - 3} d y

diverges

Solution steps

\int_{2}^{3} \frac{1}{y - 3} d y

Apply u-substitution

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

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= [ln ∣ u ∣]_{- 1}^{0}

Compute the boundaries:

diverges

= diverges

\int_{0}^{1} \frac{x e ^{x}}{1 + e ^{x}} d x

\int_{0}^{\frac{π}{3}} tan (θ) d θ

\int_{- π}^{π} (x^{2} + x) sin (x) d x

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

\int_{- 1}^{0} 5 \frac{e ^{\frac{1}{x}}}{x ^{3}} d x

What is the integral from 2 to 3 of 1/(y-3) ?
The integral from 2 to 3 of 1/(y-3) is diverges