How do you integrate $\int ln (e^{2 x - 1}) d x$ $int ln(e^(2x-1))dx$ ?

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Answer 1 · 2018-06-26T08:59:58

$x^{2} - x + c$ $x^2 - x +c$

Use knowledge of logarithms

${log}_{a} a^{f (x)} \equiv f (x) {ln}_{a} a \equiv f (x)$ $log_a a^f(x) -= f(x) ln_a a -= f(x)$

Hence ${ln e}^{f (x)} \equiv f (x)$ $ln e^f(x) -= f(x)$

$\Rightarrow \int ln (e^{2 x - 1}) d x = \int (2 x - 1) d x$ $=> int ln (e^(2x-1) ) dx = int (2x-1) dx$

Use the reverse power rule:

$= x^{2} - x + c$ $= x^2 -x + c$

How do you integrate int ln(e^(2x-1))dx∫ln(e2x−1)dxint ln(e^(2x-1))dx?