What is the integral of ln(u+1) ?

The integral of ln(u+1) is uln(u+1)+ln(u+1)-u-1+C

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integral of ln(u+1)

\int ln (u + 1) d u

u ln (u + 1) + ln (u + 1) - u - 1 + C

Solution steps

\int ln (u + 1) d u

Apply u-substitution

: \int ln (v) d v

= \int ln (v) d v

Apply Integration By Parts

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

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

\int 1 d v = v

= v ln (v) - v

Substitute back

v = u + 1

= (u + 1) ln (u + 1) - (u + 1)

Simplify

(u + 1) ln (u + 1) - (u + 1) : u ln (u + 1) + ln (u + 1) - u - 1

= u ln (u + 1) + ln (u + 1) - u - 1

Add a constant to the solution

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

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What is the integral of ln(u+1) ?
The integral of ln(u+1) is uln(u+1)+ln(u+1)-u-1+C