<s>[INST] why isn't $\int \ln\left((x+1)(x+2)\right)\ dx = 1/(x+1) + 1/(x+2) + C$? by the product rule for Logarithm, 
$$\ln(xy) = \ln(x) + \ln(y) $$
the question is, $\int \ln\left(x^2 + 3x + 2\right)\ dx = \int \ln\left((x+1)(x+2)\right)\ dx$
so, I thought I could divide them into $\int\ln(x+1)\ dx + \int\ln(x+2)\ dx$ and the answer will be $\1/(x+1) + 1(x+2) + C$, 
however, the system gave me that that is wrong answer and system showed another answer with using another formular. 
So I want to know why my answer is wrong. 
Thank you! 
 [/INST] This is because the derivative of the natural logarithm is the reciprocal function. The antiderivative of the natural logarithm function is:
$$ \int \ln x \, dx = x \ln x - x + C $$
So:
\begin{align*}
\int \ln(x^2 + 3x + 2) \, dx &= \int \ln[(x+1)(x+2)] \, dx \\
&= \int \ln[(x+1)(x+2)] \, dx \\
&= \int [\ln(x+1) + \ln(x+2)] \, dx \\
&= (x+1)\ln(x+1) - (x+1) + (x+2)\ln(x+2) - (x+2) + C \\[5pt]
&= (x+1)\ln(x+1) + (x+2)\ln(x+2) - 2x + C \\[5pt]
&= x\ln(x^2+3x+2) + \ln(x+1) + 2\ln(x+2) - 2x + C \\
\end{align*}
</s>