Question

Find the indefinite integral. $$ \int \ln \left(e^{2 x-1}\right) d x $$

Short answer

The indefinite integral of \(\int \ln \left(e^{2x-1}\right) dx\) is \(x^2 - x + C\).

Step-by-step solution

Simplify Logarithmic Expression

Start by using the property of logarithms that \(\ln a^b = b \ln a\), to simplify the function inside the integral: \[\int \ln \left(e^{2x-1}\right) dx = \int (2x-1) \ln (e) dx\] Since \(\ln (e) = 1\), the integrand simplifies to \((2x - 1)\). Thus the integral is: \[\int (2x - 1) dx \]

Apply Power Rule for Integration

Apply the power rule for integration which states that the integral of \(x^n\) is \(\frac{x^{n+1}}{n+1}\). The rule is applied to both terms in the integrand separately. The power rule is applied to \(2x\) treating \(2x\) as \(2*x^1\) and to '-1' treating it as \(-1*x^0\). \[ \int (2x - 1) dx = \int 2x dx - \int dx = x^2 - x \]

Add Constant of Integration

Finally, when finding an indefinite integral, always add a constant of integration, typically denoted as 'C'. Thus, the solution to the integral is: \[x^2 - x + C\]