\( \int\frac{2x+1}{x^{2}+x+2}dx\) is $$ \text{ (a) \( log(2x+1)+c \) where c is an arbitrary constant } $$ $$ \text{ (b) \( log\frac{(2x+1)}{(x^{2}+x+2)}+c \)where c is an arbitrary constant } $$ $$ \text{ (c) \( log(x^{2}+x+2)+c \) where c is an arbitrary constant } $$ $$ \text{ (d) \( log(\frac{1}{2})+c \) where C c is an arbitrary constant} $$
Step-by-step Solution:
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) \(\log(x^{2}+x+2) + C\)