What is the integral of ln(e^{2t-1}) ?

The integral of ln(e^{2t-1}) is t^2-t+C

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integral of ln(e^{2t-1})

\int ln (e^{2 t - 1}) d t

t^{2} - t + C

Solution steps

\int ln (e^{2 t - 1}) d t

Simplify

ln (e^{2 t - 1}) : 2 t - 1

= \int 2 t - 1 d t

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int 2 t d t - \int 1 d t

\int 2 t d t = t^{2}

\int 1 d t = t

= t^{2} - t

Add a constant to the solution

= t^{2} - t + C

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