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

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

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integral from 1 to 2 of ln(x-1)

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

- 1

Solution steps

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

Apply u-substitution

: \int_{0}^{1} ln (u) d u

= \int_{0}^{1} ln (u) d u

Apply Integration By Parts

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

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

\int 1 d u = u

= [u ln (u) - u]_{0}^{1}

Compute the boundaries:

- 1

= - 1

f (x) = e^{x} - x

\int \frac{cos ( \frac{x}{3} )}{sin ^{4} ( \frac{x}{3} )} d x

\int e^{7 x} cos (3 x) d x

\int (sin (x))^{\frac{1}{2}} cos (x) d x

\frac{1}{4} y^{^{''}} + y^{^{'}} + y = x^{2} - 4 x

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