Evaluate the integral ∫ (x^2 + 1)/(x^2) dx.

Evaluate the integral ∫ (x^2 + 1)/(x^2) dx.

• A. ln|x| + 2/x + C
• B. 2ln|x| + C
• C. ln|x^2| + 2/x + C
• D. ln|x| + x + C

Answer: (A)

To evaluate the integral ∫ (x^2 + 1)/(x^2) dx, we start by simplifying the integrand. This can be expressed as: \[ \frac{x^2 + 1}{x^2} = \frac{x^2}{x^2} + \frac{1}{x^2} = 1 + \frac{1}{x^2}. \] Now we can separate the integral into two simpler parts: \[ \int \left(1 + \frac{1}{x^2}\right) dx = \int 1 \, dx + \int \frac{1}{x^2} \, dx. \] The first integral, ∫ 1 dx, is straightforward: \[ \int 1 \, dx = x. \] The second integral, ∫ (1/x^2) dx, can be rewritten using the power rule. Recall that 1/x^2 can be expressed as x^(-2). Thus: \[ \int \frac{1}{x^2} \, dx = \int x^{-2} \, dx = \frac{x^{-1}}{-1} = -\frac

To evaluate the integral ∫ (x^2 + 1)/(x^2) dx, we start by simplifying the integrand. This can be expressed as:

[

\frac{x^2 + 1}{x^2} = \frac{x^2}{x^2} + \frac{1}{x^2} = 1 + \frac{1}{x^2}.

]

Now we can separate the integral into two simpler parts:

[

\int \left(1 + \frac{1}{x^2}\right) dx = \int 1 , dx + \int \frac{1}{x^2} , dx.

]

The first integral, ∫ 1 dx, is straightforward:

[

\int 1 , dx = x.

]

The second integral, ∫ (1/x^2) dx, can be rewritten using the power rule. Recall that 1/x^2 can be expressed as x^(-2). Thus:

[

\int \frac{1}{x^2} , dx = \int x^{-2} , dx = \frac{x^{-1}}{-1} = -\frac