<s>[INST] Calculating the integral $\int \frac{14x+1}{2-7x-4x^2} \, dx$ We have the following integral
$$
\int \frac{14x+1}{2-7x-4x^2} \, dx
$$
By factoring the denominator we get
$$
-\int \frac{14x+1}{4x^2+7x-2} \, dx = -\int \frac{14x+1}{(4x-1)(x+2)} \, dx
$$
And breaking into partial fractions
$$
-\int \frac{2}{4x-1} + \frac{3}{x+2} \, dx = -\int \frac{2}{4x-1} \, dx - \int \frac{3}{x+2} \, dx = - 2\int \frac{1}{4x-1} \, dx - 3 \int \frac{1}{x+2} \, dx
$$
Then by substituting (and resubstituting) we get
$$
- 2\int \frac{1}{4u} \, du - 3 \int \frac{1}{v} \, dv = -\frac{2}{4} \ln(u)-3 \ln (v)=-\frac{1}{2}\ln(4x-1)-3\ln (x+2)
$$
However, the expected result is
$$
-\frac{1}{2}\ln(1-4x)-3\ln (x+2)
$$
Have I made a mistake somewhere? If so, where?
 [/INST] Too long for a comment:
@Bernkastel is absolutely (haha) right. Your solution $+c$ is valid on a certain (open) portion of the real line, and the advertised solution on an other.  With absolute values thrown in, one gets the largest domain for a solution over the reals.
To be more explicit: on the half-line $x<0$, $ |x| = -x$. So, calculating the derivative on that interval, one gets (by the chain rule)
$$ \left(\ln |x|\right)' = \left(\ln \left(-x\right) \right)' = {1\over -x} \cdot (-1) = {1\over x};$$
similarly on the half-line $x>0$.
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