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

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

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

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

2 x ln (x + 1) - 2 (x + 1 - ln ∣ x + 1 ∣) + C

Solution steps

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

Apply Integration By Parts

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

\int \frac{2 x}{x + 1} d x = 2 (x + 1 - ln ∣ x + 1 ∣)

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

Apply log rule

lo g_{a} (x^{b}) = b \cdot lo g_{a} (x),

assuming

x \geq 0

= 2 x ln (x + 1) - 2 (x + 1 - ln ∣ x + 1 ∣)

Add a constant to the solution

= 2 x ln (x + 1) - 2 (x + 1 - ln ∣ x + 1 ∣) + C

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