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

The integral of (ln(2x+1))/(2x+1) is 1/4 ln^2(2x+1)+C

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integral of (ln(2x+1))/(2x+1)

\int \frac{ln ( 2 x + 1 )}{2 x + 1} d x

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

Solution steps

\int \frac{ln ( 2 x + 1 )}{2 x + 1} d x

Apply u-substitution

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

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

Apply u-substitution

: \int v d v

= \frac{1}{2} \cdot \int v d v

Apply the Power Rule

: \frac{v ^{2}}{2}

= \frac{1}{2} \cdot \frac{v ^{2}}{2}

Substitute back

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

Simplify

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

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

Add a constant to the solution

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

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What is the integral of (ln(2x+1))/(2x+1) ?
The integral of (ln(2x+1))/(2x+1) is 1/4 ln^2(2x+1)+C