Question

$\int \frac{\log x - \log^2 x + x^2}{x^3} dx$ is:

• A. $\frac{\log x + 2x \log x + C}{2x^2}$
• B. $\log^2 x + 2x \log x + C$
• C. $\frac{\log^2 x + 2x^2 \log x + C}{2x^2}$
• D. $\frac{\log x + 2x^2 \log x + C}{2x^2}$

Answer 1

Answer: (C)

$\frac{\log^2 x + 2x^2 \log x + C}{2x^2}$

Answer 2

The integral to be evaluated is $I = \int \frac{\log x - \log^2 x + x^2}{x^3} dx$.

We can split the integrand into two parts: $I = \int \left( \frac{\log x - \log^2 x}{x^3} + \frac{x^2}{x^3} \right) dx$ $I = \int \frac{\log x - \log^2 x}{x^3} dx + \int \frac{1}{x} dx$

Let's evaluate the second integral first: $\int \frac{1}{x} dx = \log |x| + C_1$.

Now, let's consider the first integral: $\int \frac{\log x - \log^2 x}{x^3} dx$. Consider the function $f(x) = \frac{\log^2 x}{x^2}$. Using the quotient rule for differentiation, $\frac{d}{dx} \left( \frac{u}{v} \right) = \frac{u'v - uv'}{v^2}$: Here, $u = \log^2 x$ and $v = x^2$. $u' = \frac{d}{dx}(\log^2 x) = 2 \log x \cdot \frac{1}{x}$ (using chain rule). $v' = \frac{d}{dx}(x^2) = 2x$.

So, $\frac{d}{dx} \left( \frac{\log^2 x}{x^2} \right) = \frac{\left(2 \log x \cdot \frac{1}{x}\right) \cdot x^2 - (\log^2 x) \cdot (2x)}{(x^2)^2}$ $= \frac{2x \log x - 2x \log^2 x}{x^4}$ $= \frac{2x (\log x - \log^2 x)}{x^4}$ $= \frac{2(\log x - \log^2 x)}{x^3}$.

Therefore, we have found that $\frac{d}{dx} \left( \frac{\log^2 x}{x^2} \right) = \frac{2(\log x - \log^2 x)}{x^3}$. This implies that $\frac{d}{dx} \left( \frac{1}{2} \frac{\log^2 x}{x^2} \right) = \frac{\log x - \log^2 x}{x^3}$.

So, the first integral is: $\int \frac{\log x - \log^2 x}{x^3} dx = \frac{1}{2} \frac{\log^2 x}{x^2} + C_2$.

Now, combine the results of both integrals: $I = \left( \frac{1}{2} \frac{\log^2 x}{x^2} \right) + \log x + C$, where $C = C_1 + C_2$. To match the given options, we can write the expression with a common denominator $2x^2$: $I = \frac{\log^2 x}{2x^2} + \frac{2x^2 \log x}{2x^2} + C$ $I = \frac{\log^2 x + 2x^2 \log x}{2x^2} + C$.