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

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

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integral from 1 to 3 of ln(x-1)

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

2 ln (2) - 2

Decimal

- 0.61370 \dots

Solution steps

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

Apply u-substitution

: \int_{0}^{2} ln (u) d u

= \int_{0}^{2} ln (u) d u

Apply Integration By Parts

: [u ln (u) - \int 1 d u]_{0}^{2}

= [u ln (u) - \int 1 d u]_{0}^{2}

\int 1 d u = u

= [u ln (u) - u]_{0}^{2}

Compute the boundaries:

2 ln (2) - 2

= 2 ln (2) - 2

n = 1 \sum \infty cos (2 n)

y = x^{2} - 1, y = x + 2, x = 0, x + 1

x \to - 1 lim (5 - 3 x - x^{2})

(3.1) (0)

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

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