What is the integral from 4 to 8 of 1/(x-3) ?

The integral from 4 to 8 of 1/(x-3) is ln(5)

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integral from 4 to 8 of 1/(x-3)

\int_{4}^{8} \frac{1}{x - 3} d x

ln (5)

Solution steps

\int_{4}^{8} \frac{1}{x - 3} d x

Apply u-substitution

: \int_{1}^{5} \frac{1}{u} d u

= \int_{1}^{5} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= [ln ∣ u ∣]_{1}^{5}

Compute the boundaries:

ln (5)

= ln (5)

\int_{- 1}^{3} 5 x x + 1 d x

\int_{0}^{x} (1 - \frac{r}{x})^{\frac{1}{n}} r d r

\int_{5}^{12} x d x

\int_{1}^{2} x^{5} e^{x^{2}} d x

\int_{0}^{1} 4 arctan (x^{2}) d x

What is the integral from 4 to 8 of 1/(x-3) ?
The integral from 4 to 8 of 1/(x-3) is ln(5)