What is the integral from 5 to 11 of (16)/(x-1) ?

The integral from 5 to 11 of (16)/(x-1) is 16ln(5/2)

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integral from 5 to 11 of (16)/(x-1)

\int_{5}^{11} \frac{16}{x - 1} d x

16 ln (\frac{5}{2})

Decimal Notation

14.66065 \dots

Solution steps

\int_{5}^{11} \frac{16}{x - 1} d x

Take the constant out:

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

= 16 \cdot \int_{5}^{11} \frac{1}{x - 1} d x

Apply u-substitution

: \int_{4}^{10} \frac{1}{u} d u

= 16 \cdot \int_{4}^{10} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 16 [ln ∣ u ∣]_{4}^{10}

Compute the boundaries:

ln (10) - 2 ln (2)

= 16 (ln (10) - 2 ln (2))

Simplify

= 16 ln (\frac{5}{2})

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

\frac{d}{d x} (\frac{tan ( x )}{x})

n = 0 \sum \infty \frac{n}{3 ^{n}}

f (x) = arccos (2 x)

x \to 0 + lim (\frac{e ^{\frac{1}{x}}}{x})

What is the integral from 5 to 11 of (16)/(x-1) ?
The integral from 5 to 11 of (16)/(x-1) is 16ln(5/2)