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

The integral of ln(y+1) is yln(y+1)+ln(y+1)-y-1+C

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integral of ln(y+1)

\int ln (y + 1) d y

y ln (y + 1) + ln (y + 1) - y - 1 + C

Solution steps

\int ln (y + 1) d y

Apply u-substitution

= \int ln (u) d u

Apply Integration By Parts

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

\int 1 d u = u

= u ln (u) - u

Substitute back

u = y + 1

= (y + 1) ln (y + 1) - (y + 1)

Simplify

(y + 1) ln (y + 1) - (y + 1) : y ln (y + 1) + ln (y + 1) - y - 1

= y ln (y + 1) + ln (y + 1) - y - 1

Add a constant to the solution

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

What is the integral of ln(y+1) ?
The integral of ln(y+1) is yln(y+1)+ln(y+1)-y-1+C