What is the integral of (ln(x+1))^2 ?

The integral of (ln(x+1))^2 is ln^2(x+1)(x+1)-2(xln(x+1)+ln(x+1)-x-1)+C

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integral of (ln(x+1))^2

\int (ln (x + 1))^{2} d x

ln^{2} (x + 1) (x + 1) - 2 (x ln (x + 1) + ln (x + 1) - x - 1) + C

Solution steps

\int (ln (x + 1))^{2} d x

Simplify

= \int ln^{2} (x + 1) d x

Apply u-substitution

= \int ln^{2} (u) d u

Apply Integration By Parts

= u ln^{2} (u) - \int 2 ln (u) d u

\int 2 ln (u) d u = 2 (u ln (u) - u)

= u ln^{2} (u) - 2 (u ln (u) - u)

Substitute back

u = x + 1

= (x + 1) ln^{2} (x + 1) - 2 ((x + 1) ln (x + 1) - (x + 1))

Expand

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

= ln^{2} (x + 1) (x + 1) - 2 (x ln (x + 1) + ln (x + 1) - x - 1)

Add a constant to the solution

= ln^{2} (x + 1) (x + 1) - 2 (x ln (x + 1) + ln (x + 1) - x - 1) + C

What is the integral of (ln(x+1))^2 ?
The integral of (ln(x+1))^2 is ln^2(x+1)(x+1)-2(xln(x+1)+ln(x+1)-x-1)+C