What is the integral of the function (x^2 + 1)/x dx?

What is the integral of the function (x^2 + 1)/x dx?

• A. x + ln|x| + C
• B. ln|x^2 + 1| + C
• C. (1/2)x^2 + C
• D. x^2 + C

Answer: (A)

To solve the integral of the function \((x^2 + 1)/x\) dx, we can first simplify the integrand by breaking it down into simpler fractions: \[ \frac{x^2 + 1}{x} = \frac{x^2}{x} + \frac{1}{x} = x + \frac{1}{x} \] Now we can integrate the simplified expression: \[ \int \left( x + \frac{1}{x} \right) dx \] The integral of \(x\) is straightforward: \[ \int x \, dx = \frac{1}{2} x^2 \] For \(\frac{1}{x}\), the integral is: \[ \int \frac{1}{x} \, dx = \ln |x| \] Combining these results gives us: \[ \int (x + \frac{1}{x}) \, dx = \frac{1}{2} x^2 + \ln |x| + C \] It seems there might be some discrepancy, as the answer provided is x + ln|x| + C. However, if we differentiate

To solve the integral of the function ((x^2 + 1)/x) dx, we can first simplify the integrand by breaking it down into simpler fractions:

[

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

]

Now we can integrate the simplified expression:

[

\int \left( x + \frac{1}{x} \right) dx

]

The integral of (x) is straightforward:

[

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

]

For (\frac{1}{x}), the integral is:

[

\int \frac{1}{x} , dx = \ln |x|

]

Combining these results gives us:

[

\int (x + \frac{1}{x}) , dx = \frac{1}{2} x^2 + \ln |x| + C

]

It seems there might be some discrepancy, as the answer provided is x + ln|x| + C. However, if we differentiate