What is the integral of log_{e}(x+1) ?

The integral of log_{e}(x+1) is xln(x+1)+ln(x+1)-x-1+C

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integral of log_{e}(x+1)

\int lo g_{e} (x + 1) d x

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

Solution steps

\int lo g_{e} (x + 1) d x

Apply log rule:

lo g_{e} (x) = ln (x)

= \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

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