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

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

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

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

x^{2} - x + C

Solution steps

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

Simplify

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

= \int 2 x - 1 d x

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 x d x - \int 1 d x

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

\int 1 d x = x

= x^{2} - x

Add a constant to the solution

= x^{2} - x + C

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What is the integral of ln(e^{(2x-1)}) ?
The integral of ln(e^{(2x-1)}) is x^2-x+C