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

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

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

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

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

Solution steps

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

Apply Integration By Parts

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

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

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

Apply log rule

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

assuming

x \geq 0

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

Add a constant to the solution

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

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