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

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

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integral of-ln(x+1)

\int - ln (x + 1) d x

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

Solution steps

\int - ln (x + 1) d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d 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

\frac{d}{d x} (\frac{b}{cos ( θ )})

\int + 17 cot^{4} (x) d x

\frac{3 d y}{d x} + 15 y = 90, y (0) = 0

\int (2 sin (x) - \frac{π}{4} x) d x

\frac{d}{d x} (ln (\frac{x}{x}))

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