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

The integral of ln(x-1) is xln(x-1)-ln(x-1)-x+1+C

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integral of ln(x-1)

\int ln (x - 1) d x

x ln (x - 1) - ln (x - 1) - x + 1 + C

Solution steps

\int ln (x - 1) d x

Apply u-substitution

: \int ln (u) d u

= \int ln (u) d u

Apply Integration By Parts

: u ln (u) - \int 1 d u

= u ln (u) - \int 1 d u

\int 1 d u = u

= u ln (u) - u

Substitute back

u = x - 1

= (x - 1) ln (x - 1) - (x - 1)

Simplify

(x - 1) ln (x - 1) - (x - 1) : x ln (x - 1) - ln (x - 1) - x + 1

= x ln (x - 1) - ln (x - 1) - x + 1

Add a constant to the solution

= x ln (x - 1) - ln (x - 1) - x + 1 + C

\int \frac{1}{x ^{2} 81 x ^{2} - 49} d x

\frac{d}{d x} (4 x^{π})

\frac{d}{d x} (3 cos^{3} (x))

\frac{d}{d x} ((1 + cos (x))^{8})

f (x) = x^{2} + 7

What is the integral of ln(x-1) ?
The integral of ln(x-1) is xln(x-1)-ln(x-1)-x+1+C