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

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

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integral of 2ln(x+1)

\int 2 ln (x + 1) d x

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

Solution steps

\int 2 ln (x + 1) d x

Take the constant out:

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

= 2 \cdot \int ln (x + 1) d x

Apply u-substitution

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

Apply Integration By Parts

= 2 (u ln (u) - \int 1 d u)

\int 1 d u = u

= 2 (u ln (u) - u)

Substitute back

u = 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

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

Add a constant to the solution

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

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