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

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

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integral of ln(1+2x)

\int ln (1 + 2 x) d x

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

Solution steps

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

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

Simplify

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

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

Add a constant to the solution

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

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

9 sin (4 x)

w e^{w}

\frac{\partial}{\partial x} (\frac{x ^{\frac{5}{7}}}{e ^{x}})

x \to 0 lim (\frac{2}{x ^{2} + x} - \frac{2}{x})

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