<s>[INST] Indefinite integral evaluation of fraction Here's the indefinite integral to evaluate:
$F(x) = \int \left(\frac{3x + 4}{x + 3}\right)dx$
Here are my steps (verbose):
$$\begin{align}
u & = x + 3 \\
(x + 3)dx & = du \\
\left(\frac{d}{dx}(x + 3)\right)dx & = du \\
\left(\frac{d}{dx}(x) + \frac{d}{dx}(3)\right)dx & = du \\
\left(\frac{d}{dx}(x) + 0\right)dx & = du \\
\left( \frac{dx}{dx} + 0 \right)dx & = du \\
(1 + 0)dx & = du \\
(1)dx & = du \\
dx & = \left( \frac{1}{1}\right)du \\
dx & = du \\
F(\ ) & = \int \left( \frac{(3x + 4)}{u} \right)du \\
u & = x + 3 \\
x + 3 & = u \\
x & = u - 3 \\
F(u)& = \int \left( \frac{3(u - 3) + 4}{u} \right)du \\
& =  \int \left( \frac{3u - 9 + 4}{u} \right)du \\
& = \int \left( \frac{3u - 5}{u} \right)du \\
& = \int \left( \frac{3u}{u} - \frac{5}{u} \right)du \\
& = \int \left( 3 - \frac{5}{u} \right)du \\
& = \int (3)du - \int \left( \frac{5}{u} \right)du \\
& = \int (3)du - 5*\int \left( \frac{1}{u} \right)du \\
& = \int (3)du - 5*ln \lvert u \rvert \\
& = \int (3*u^0)du - 5*ln \lvert u \rvert \\
& = 3*\int (u^0)du - 5*ln \lvert u \rvert \\
& = 3* \frac{1}{0 + 1}*u^{0 + 1} - 5*ln \lvert u∣ \\
& = 3* \frac{1}{1}*u^1 - 5*ln \lvert u \rvert \\
& = 3u - 5*ln \lvert u \rvert \\
u & = x + 3 \\
F(x) & = 3(x + 3) - 5*ln\lvert x + 3\rvert \\
& = 3x + 9 - 5*ln\lvert x + 3\rvert \\
& = 3x - 5*ln\lvert x + 3\rvert + 9 \\
& = \bf [3x - 5*ln\lvert x + 3\rvert + 9 + C] \\
\end{align}$$
This result agrees with both symbolab and wolfram.
Here's the solution approach provided:
Since the degree of the numerator is not less than the degree of the denominator, we must divide. A clever way to do this is shown below.
$$\begin{align}
\int \left( \frac{3x+4}{x+3} \right)dx &= \int \left( \frac{3x+9−5}{x+3} \right)dx \\
&= \int \left( \frac{3(x+3)}{x+3}  − \frac{5}{x+3} \right)dx \\
&= \int \left( 3 − \frac{5}{x+3} \right)dx \\
\end{align}$$
Now the integration is straightforward:
$\int \left(3 −​\frac{5}{x+3} \right)dx = 3x−5*ln\lvert x + 3\rvert +C$ 
#
Although the partial fraction solution appears to be mathematically sound, nonetheless it loses real information: "+9"
Furthermore, I realize that if I were to take the derivative of either solution, I would end up with the same results as both constants "+9" and "C" would resolve to 0. 
That stated, my concern is that using a method that is inherently prone to losing information is dangerous as it could result in actual computational error especially if instead of an indefinite integral, one was expected to evaluate a definite integral as in determining the actual area under the functions curve, in which case the missing "+9" would result in an erroneous outcome.
Would my argument above justify contesting the method provided?
 [/INST] Nice work!
About your misgivings: Evaluating the definite integral  $F(x) \Big|_b^a$ will yield $$ 3x - 5\ln\lvert x + 3\rvert + 9 + C\Big|_b^a = \Big(3a - 5\ln|a+3| + 9 +C\Big)  - \Big( 3b - 5\ln |b + 3| + 9 + C \Big)$$
$$= 3a - 5\ln|a+3| +9 +C - 3b + 5\ln |b + 3| - 9 - C]$$
You can see that  upon simplification, we encounter $9 - 9$, $C-C$.
Constants of integration all cancel in the subtraction $F(a) - F(b)$, and as a result, they play no roll in definite integration.
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