Let's solve these integrals step-by-step.

Problem 47:

The integral to evaluate is: $\int \frac{\ln x}{x^2} \, dx$

Let's solve this by using integration by parts. Set:

1. $u = \ln x \Rightarrow du = \frac{1}{x} \, dx$
2. $dv = \frac{1}{x^2} \, dx = x^{-2} \, dx \Rightarrow v = -\frac{1}{x}$

Applying integration by parts, $\int u \, dv = uv - \int v \, du$:

$\int \frac{\ln x}{x^2} \, dx = -\frac{\ln x}{x} - \int -\frac{1}{x} \cdot \frac{1}{x} \, dx$ $= -\frac{\ln x}{x} + \int \frac{1}{x^2} \, dx$ $= -\frac{\ln x}{x} - \frac{1}{x} + C$

So, the result for this integral is: $\int \frac{\ln x}{x^2} \, dx = -\frac{\ln x}{x} - \frac{1}{x} + C$

Problem 48:

The definite integral to evaluate is: $\int_1^e x^2 \ln x \, dx$

We can solve this using integration by parts as well. Set:

1. $u = \ln x \Rightarrow du = \frac{1}{x} \, dx$
2. $dv = x^2 \, dx \Rightarrow v = \frac{x^3}{3}$

Then, applying integration by parts: $\int x^2 \ln x \, dx = \frac{x^3}{3} \ln x - \int \frac{x^3}{3} \cdot \frac{1}{x} \, dx$ $= \frac{x^3}{3} \ln x - \int \frac{x^2}{3} \, dx$ $= \frac{x^3}{3} \ln x - \frac{x^3}{9} + C$

Now, we evaluate this expression from $x = 1$ to $x = e$: $\int_1^e x^2 \ln x \, dx = \left( \frac{e^3}{3} \ln e - \frac{e^3}{9} \right) - \left( \frac{1^3}{3} \ln 1 - \frac{1^3}{9} \right)$

Since $\ln e = 1$ and $\ln 1 = 0$, this simplifies to: $= \left( \frac{e^3}{3} - \frac{e^3}{9} \right) - \left( 0 - \frac{1}{9} \right)$ $= \frac{2e^3}{9} + \frac{1}{9}$ $= \frac{2e^3 + 1}{9}$

So, the result for this definite integral is: $\int_1^e x^2 \ln x \, dx = \frac{2e^3 + 1}{9}$

Would you like more details on these steps or further clarification?

Here are five related questions to deepen your understanding:

1. How would you apply integration by parts to different types of functions?
2. What are alternative techniques for evaluating integrals involving logarithmic functions?
3. How can definite integrals be applied to real-world problems?
4. Why is it helpful to recognize when integration by parts is appropriate?
5. What is the significance of constants in indefinite integrals?

Tip: When integrating by parts, remember to choose $u$ and $dv$ such that $dv$ is easy to integrate, and $du$ simplifies the integral.