Find

1. Rewrite the integral so the numerator is the exact derivative of the denominator.

$\int \frac{3e^x}{5 - 0.3e^x} \, dx = -10 \int \frac{-0.3e^x}{5 - 0.3e^x} \, dx$

2. Integrate using the standard form $\int \frac{f'(x)}{f(x)} \, dx = \ln|f(x)| + c$.

$= -10 \ln|5 - 0.3e^x| + c$

1. Apply integration by parts to the given integral.

$\int \cos(\ln x) \, dx$
Let $u = \cos(\ln x)$ and $\frac{dv}{dx} = 1$.
Then using the chain rule, $\frac{du}{dx} = -\sin(\ln x) \cdot \frac{1}{x}$, and $v = x$.
Using the integration by parts formula $\int u \frac{dv}{dx} \, dx = uv - \int v \frac{du}{dx} \, dx$:
$\int \cos(\ln x) \, dx = x \cos(\ln x) - \int x \left( -\frac{\sin(\ln x)}{x} \right) \, dx$
$= x \cos(\ln x) + \int \sin(\ln x) \, dx$

2. Apply integration by parts a second time to the new integral.

For $\int \sin(\ln x) \, dx$, let $u = \sin(\ln x)$ and $\frac{dv}{dx} = 1$.
Then $\frac{du}{dx} = \cos(\ln x) \cdot \frac{1}{x}$ and $v = x$.
Applying integration by parts again:
$\int \sin(\ln x) \, dx = x \sin(\ln x) - \int x \left( \frac{\cos(\ln x)}{x} \right) \, dx$
$= x \sin(\ln x) - \int \cos(\ln x) \, dx$

3. Solve algebraically for the original integral.

Let $I = \int \cos(\ln x) \, dx$.
Substituting the result from the second integration by parts into the first equation:
$I = x \cos(\ln x) + \left( x \sin(\ln x) - I \right)$
$2I = x \cos(\ln x) + x \sin(\ln x)$
$I = \frac{1}{2} x (\cos(\ln x) + \sin(\ln x)) + C$
where $C$ is an arbitrary constant of integration.

1. Determine the root of the expression inside the modulus to split the integral.

$2e^x - 5 = 0$
$e^x = \frac{5}{2}$
$x = \ln\left(\frac{5}{2}\right)$

2. Set up the definite integrals using the definition of absolute value over the intervals.

For $0 \leq x < \ln\left(\frac{5}{2}\right)$, $2e^x - 5 < 0$.
For $\ln\left(\frac{5}{2}\right) \leq x \leq 3$, $2e^x - 5 \geq 0$.
$\int_0^3 |2e^x - 5| \, dx = \int_0^{\ln(5/2)} -(2e^x - 5) \, dx + \int_{\ln(5/2)}^3 (2e^x - 5) \, dx$

3. Integrate the expressions and substitute the limits.

$= \left[ 5x - 2e^x \right]_0^{\ln(5/2)} + \left[ 2e^x - 5x \right]_{\ln(5/2)}^3$
$= \left( 5\ln\left(\frac{5}{2}\right) - 2e^{\ln(5/2)} \right) - (0 - 2e^0) + \left( 2e^3 - 15 \right) - \left( 2e^{\ln(5/2)} - 5\ln\left(\frac{5}{2}\right) \right)$

4. Simplify to obtain the exact value.

Since $e^{\ln(5/2)} = \frac{5}{2}$ and $e^0 = 1$:
$= \left( 5\ln\left(\frac{5}{2}\right) - 2\left(\frac{5}{2}\right) \right) - (-2) + \left( 2e^3 - 15 \right) - \left( 2\left(\frac{5}{2}\right) - 5\ln\left(\frac{5}{2}\right) \right)$
$= \left( 5\ln\left(\frac{5}{2}\right) - 5 + 2 \right) + \left( 2e^3 - 15 - 5 + 5\ln\left(\frac{5}{2}\right) \right)$
$= 5\ln\left(\frac{5}{2}\right) - 3 + 2e^3 - 20 + 5\ln\left(\frac{5}{2}\right)$
$= 2e^3 + 10\ln\left(\frac{5}{2}\right) - 23$