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

The integral of ln(2t+1) is tln(2t+1)+1/2 ln(2t+1)-t-1/2+C

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integral of ln(2t+1)

\int ln (2 t + 1) d t

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

Solution steps

\int ln (2 t + 1) d t

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 t + 1

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

Simplify

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

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

Add a constant to the solution

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

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