You have provided the following integral:

$\int_1^e \frac{x^2 + 1}{x^3} \, dx$

To solve this, we will first simplify the integrand:

$\frac{x^2 + 1}{x^3} = \frac{x^2}{x^3} + \frac{1}{x^3} = \frac{1}{x} + \frac{1}{x^3}$

So, the integral becomes:

$\int_1^e \left( \frac{1}{x} + \frac{1}{x^3} \right) \, dx$

Now, let's integrate each term separately:

1. The integral of $\frac{1}{x}$ is $\ln|x|$.
2. The integral of $\frac{1}{x^3}$ is $-\frac{1}{2x^2}$.

Thus, the integral becomes:

$\int_1^e \left( \frac{1}{x} + \frac{1}{x^3} \right) \, dx = \left[ \ln|x| - \frac{1}{2x^2} \right]_1^e$

Evaluating at the limits:

• When $x = e$:

$\ln|e| - \frac{1}{2e^2} = \ln(e) - \frac{1}{2e^2} = 1 - \frac{1}{2e^2}$

• When $x = 1$:

$\ln|1| - \frac{1}{2(1)^2} = 0 - \frac{1}{2} = -\frac{1}{2}$

Now, subtract the lower limit evaluation from the upper limit:

$\left( 1 - \frac{1}{2e^2} \right) - \left( -\frac{1}{2} \right) = 1 - \frac{1}{2e^2} + \frac{1}{2}$

Simplifying:

$1 + \frac{1}{2} - \frac{1}{2e^2} = \frac{3}{2} - \frac{1}{2e^2}$

Thus, the value of the integral is:

$\boxed{\frac{3}{2} - \frac{1}{2e^2}}$

Would you like further clarification or a breakdown of any specific steps? Here are some related questions you could explore:

1. How do we integrate rational functions like $\frac{1}{x^3}$?
2. What are the common techniques for handling logarithmic integrals?
3. How does the substitution method apply to this integral?
4. What are the properties of the natural logarithm function in integrals?
5. How do we evaluate definite integrals involving exponential functions like $e^x$?

Tip: When you encounter fractions involving powers of $x$, it's often helpful to break them into simpler fractions, like $\frac{x^2}{x^3} = \frac{1}{x}$, before integrating.