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

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

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integral of 2xln(x+1)

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

x^{2} ln (x + 1) + 2 x - ln (x + 1) - \frac{1}{2} (x + 1)^{2} + 2 + C

Solution steps

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

Apply u-substitution

: \int (u - 1) ln (u) d u

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

Apply Integration By Parts

: ln (u) (\frac{u ^{2}}{2} - u) - \int \frac{1}{2} (u - 2) d u

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

\int \frac{1}{2} (u - 2) d u = \frac{u ^{2}}{4} - u

= 2 (ln (u) (\frac{u ^{2}}{2} - u) - (\frac{u ^{2}}{4} - u))

Substitute back

u = x + 1

= 2 (ln (x + 1) (\frac{( x + 1 ) ^{2}}{2} - (x + 1)) - (\frac{( x + 1 ) ^{2}}{4} - (x + 1)))

Simplify

2 (ln (x + 1) (\frac{( x + 1 ) ^{2}}{2} - (x + 1)) - (\frac{( x + 1 ) ^{2}}{4} - (x + 1))) : x^{2} ln (x + 1) + 2 x - ln (x + 1) - \frac{1}{2} (x + 1)^{2} + 2

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

Add a constant to the solution

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

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