What is the integral of ln|x+1| ?

The integral of ln|x+1| is xln|x+1|-x-1+ln|x+1|+C

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integral of ln|x+1|

\int ln ∣ x + 1 ∣ d x

x ln ∣ x + 1 ∣ - x - 1 + ln ∣ x + 1 ∣ + C

Solution steps

\int ln ∣ x + 1 ∣ d x

Apply Integration By Parts

: x ln ∣ x + 1 ∣ - \int \frac{x}{x + 1} d x

= x ln ∣ x + 1 ∣ - \int \frac{x}{x + 1} d x

\int \frac{x}{x + 1} d x = x + 1 - ln ∣ x + 1 ∣

= x ln ∣ x + 1 ∣ - (x + 1 - ln ∣ x + 1 ∣)

Simplify

= x ln ∣ x + 1 ∣ - x - 1 + ln ∣ x + 1 ∣

Add a constant to the solution

= x ln ∣ x + 1 ∣ - x - 1 + ln ∣ x + 1 ∣ + C

\frac{d}{d x} ((x^{2} - 3 x)^{5})

(12 x^{2} + y - 1) d x - (6 y - x) d y = 0, y (1) = 0

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

\frac{d y}{d x}, y = 5^{x}

\frac{d}{d x} (16 x^{2} + 9)

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