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

The integral of xln(2x+1) is 1/16 (8x^2ln(2x+1)-4x^2+4x-2ln(2x+1)+3)+C

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integral of xln(2x+1)

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

\frac{1}{16} (8 x^{2} ln (2 x + 1) - 4 x^{2} + 4 x - 2 ln (2 x + 1) + 3) + C

Solution steps

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

Apply u-substitution

: \int \frac{ln ( u ) ( u - 1 )}{4} d u

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

Take the constant out:

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

= \frac{1}{4} \cdot \int ln (u) (u - 1) d u

Apply Integration By Parts

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

= \frac{1}{4} (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

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

Substitute back

u = 2 x + 1

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

Simplify

\frac{1}{4} (ln (2 x + 1) (\frac{( 2 x + 1 ) ^{2}}{2} - (2 x + 1)) - (\frac{( 2 x + 1 ) ^{2}}{4} - (2 x + 1))) : \frac{1}{16} (8 x^{2} ln (2 x + 1) - 4 x^{2} + 4 x - 2 ln (2 x + 1) + 3)

= \frac{1}{16} (8 x^{2} ln (2 x + 1) - 4 x^{2} + 4 x - 2 ln (2 x + 1) + 3)

Add a constant to the solution

= \frac{1}{16} (8 x^{2} ln (2 x + 1) - 4 x^{2} + 4 x - 2 ln (2 x + 1) + 3) + C

\frac{\partial}{\partial x} (3 y^{2} + x cos (x y))

\int \frac{x ^{3} + 1}{x + 1} d x

x \to 3 - lim (\frac{3}{x - 3})

\frac{d}{d x} (2 x - 2 - 2 cos (x - 2))

3 x + 2

What is the integral of xln(2x+1) ?
The integral of xln(2x+1) is 1/16 (8x^2ln(2x+1)-4x^2+4x-2ln(2x+1)+3)+C