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! 

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*}
