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

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

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integral of ln(2x+1)

\int ln (2 x + 1) d x

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

Solution steps

\int ln (2 x + 1) d x

Apply u-substitution

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

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

Take the constant out:

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

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

Apply Integration By Parts

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

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

\int 1 d u = u

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

Substitute back

u = 2 x + 1

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

Simplify

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

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

Add a constant to the solution

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

x (\frac{d y}{d x}) - y = 2 x^{2}

\int \frac{e ^{x}}{( 6 - e ^{x} ) ^{2}} d x

\frac{\partial}{\partial y} (x^{2} y^{4})

x \to 2 lim (\frac{x ^{2} - 2}{x - 2})

n = 0 \sum \infty 2 + 0.8 6^{n}

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