Question

$\int \frac{e^{4\log x}-e^{5\log x}}{x^5}dx$

Answer 1

$\log |x| - x + C$

Answer 2

To solve the integral $\int \frac{e^{4\log x}-e^{5\log x}}{x^5}dx$, we first simplify the terms involving logarithms.

Step 1: Simplify the exponential terms using logarithm properties. Recall the logarithm property: $a \log b = \log (b^a)$. So, $e^{4\log x} = e^{\log (x^4)}$. And $e^{5\log x} = e^{\log (x^5)}$.

Next, recall the property: $e^{\log y} = y$. Using this, we get: $e^{\log (x^4)} = x^4$ $e^{\log (x^5)} = x^5$

Step 2: Substitute the simplified terms back into the integral. The integral becomes: $\int \frac{x^4 - x^5}{x^5}dx$

Step 3: Simplify the integrand. Divide each term in the numerator by the denominator: $\frac{x^4 - x^5}{x^5} = \frac{x^4}{x^5} - \frac{x^5}{x^5} = \frac{1}{x} - 1$

Step 4: Integrate the simplified expression. Now, the integral is: $\int \left(\frac{1}{x} - 1\right)dx$ Integrate each term separately: $\int \frac{1}{x}dx - \int 1 dx$ Using the standard integral formulas $\int \frac{1}{x}dx = \log |x| + C$ and $\int k dx = kx + C$: $\log |x| - x + C$ where $C$ is the constant of integration.

The final answer is $\log |x| - x + C$.

Explanation of the solution: The problem involves simplifying exponential terms with logarithmic powers using the property $e^{a \log x} = e^{\log x^a} = x^a$. After simplification, the expression becomes a simple rational function which can be split into two terms. Each term is then integrated using standard integration formulas: $\int \frac{1}{x} dx = \log|x|$ and $\int 1 dx = x$.