What is the integral from 0 to 1 of 1/(x-3) ?

The integral from 0 to 1 of 1/(x-3) is ln(2)-ln(3)

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integral from 0 to 1 of 1/(x-3)

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

ln (2) - ln (3)

Solution steps

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

Apply u-substitution

: \int_{- 3}^{- 2} \frac{1}{u} d u

= \int_{- 3}^{- 2} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= [ln ∣ u ∣]_{- 3}^{- 2}

Compute the boundaries:

ln (2) - ln (3)

= ln (2) - ln (3)

y^{^{''}} - 81 y = 0, y (2) = 2, y (- 1) = - 1

\frac{d}{d x} (x + e)

\frac{1}{sin ( x ) cos ( x + b + c )}

\int_{0}^{1} arcsin (x) - sin (x) d x

\int (\frac{3 x ^{2} + 3 x + 1}{x}) d x

What is the integral from 0 to 1 of 1/(x-3) ?
The integral from 0 to 1 of 1/(x-3) is ln(2)-ln(3)