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

\frac{\partial}{\partial x} (2 x + 2 y^{2})

t a y l or x sin (x^{3})

d er i v a t i v e x^{- \frac{2}{4}} + x^{\frac{2}{4}}

\int_{0}^{1} \frac{1}{( x - 1 ) ^{\frac{2}{3}}} d x

\frac{d}{d x} ((8 x)^{\frac{1}{3}})