Question:

Evaluate the integral: \[ \int \frac{2x+1}{x^{2}+x+2} \, dx \]

Options:

(a) \(\log(2x+1) + C\) where \(C\) is an arbitrary constant
(b) \(\log \frac{(2x+1)}{(x^{2}+x+2)} + C\) where \(C\) is an arbitrary constant
(c) \(\log(x^{2}+x+2) + C\) where \(C\) is an arbitrary constant
(d) \(\log\left(\tfrac{1}{2}\right) + C\) where \(C\) is an arbitrary constant

Solution:

We observe that the denominator is \(x^2 + x + 2\). Let: \[ f(x) = x^2 + x + 2 \] Then, \[ f'(x) = 2x + 1 \]

Notice that the numerator of the integral is exactly the derivative of the denominator: \[ \int \frac{2x+1}{x^{2}+x+2} \, dx = \int \frac{f'(x)}{f(x)} \, dx \]

This is a standard logarithmic form: \[ \int \frac{f'(x)}{f(x)} \, dx = \ln|f(x)| + C \]

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

Correct Answer:

(c) \(\log(x^{2}+x+2) + C\)