What is the integral of log_{e}(2y+1) ?

The integral of log_{e}(2y+1) is yln(2y+1)+1/2 ln(2y+1)-y-1/2+C

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integral of log_{e}(2y+1)

\int lo g_{e} (2 y + 1) d y

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

Solution steps

\int lo g_{e} (2 y + 1) d y

Apply log rule:

lo g_{e} (x) = ln (x)

= \int ln (2 y + 1) d y

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 y + 1

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

Simplify

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

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

Add a constant to the solution

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

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What is the integral of log_{e}(2y+1) ?
The integral of log_{e}(2y+1) is yln(2y+1)+1/2 ln(2y+1)-y-1/2+C