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

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

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integral of ln(1+y)

\int ln (1 + y) d y

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

Solution steps

\int ln (1 + y) d y

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 = 1 + y

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

Simplify

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

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

Add a constant to the solution

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

x^{4} - x^{2}, 1 - x^{2}

\frac{d}{d x} (cos^{3} (2 + x))

\frac{d}{d x} ((4 - x)^{\frac{1}{x - 3}})

y = ln (x + 3)

y^{^{''}} + 9 y = - sin (3 t)

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