Hello Cilian,
We're going to use partial fractions to evaluate this integral. With that said, we note that the partial fraction decomposition will take on the form
\[\frac{19x^2-x+4}{x(1+4x^2)} = \frac{A}{x}+\frac{Bx+C}{1+4x^2}\]
Multiplying both sides by the common denominator yields
\[19x^2-x+4 = A(1+4x^2) + (Bx+C)x\]
We now simplify the right hand side and group like terms to get
\[19x^2-x+4 = (4A+B)x^2+Cx+A\]
If we now compare the coefficients of both sides, we have the following system of equations:
\[\left\{\begin{aligned} 4A + B &= 19 \\ C &= -1 \\ A &= 4 \end{aligned}\right.\]
Luckily for us, we already have two of the solutions, so it follows now that $4(4)+B = 19 \implies B=3$. Therefore, we now see that
\[\frac{19x^2-x+4}{x(1+4x^2)} = \frac{4}{x} + \frac{3x-1}{1+4x^2}\]
Hence, we now see that
\[\int \frac{19x^2-x+4}{x(1+4x^2)}\,dx = \int\frac{4}{x}\,dx + \int\frac{3x-1}{1+4x^2}\,dx = \color{red}{\int\frac{4}{x}\,dx} + \color{blue}{\int\frac{3x}{1+4x^2}\,dx} - \color{green}{\int\frac{1}{1+4x^2}\,dx}\]
The first integral is rather straightforward; you should see that
\[\int\frac{4}{x}\,dx = \color{red}{4\ln|x|+C}\]
Next, for the second integral, we make a substitution: $u=1+4x^2\implies \,du =8x\,dx \implies \dfrac{du}{8}=x\,dx$. Thus,
\[\int\frac{3x}{1+4x^2}\,dx = \frac{3}{8}\int\frac{1}{u}\,du = \frac{3}{8}\ln|u|+C = \color{blue}{\frac{3}{8}\ln(1+4x^2)+C}\]
(Note here that we can drop absolute values since $1+4x^2>0$ for any $x$.)
For the last integral, we need to note that
\[\int\frac{1}{1+4x^2}\,dx = \int\frac{1}{1+(2x)^2}\,dx\]
To integrate, we make the substitution $u=2x\implies \,du = 2\,dx \implies \dfrac{du}{2}=\,dx$. Thus,
\[\int\frac{1}{1+4x^2}\,dx = \frac{1}{2}\int\frac{1}{1+u^2}\,du = \frac{1}{2}\arctan(u)+C = \color{green}{\frac{1}{2}\arctan(2x)+C}\]
Therefore, putting everything together, we have that
\[\int\frac{19x^2-x+4}{x(1+4x^2)}\,dx = 4\ln|x| + \frac{3}{8}\ln(1+4x^2) - \frac{1}{2}\arctan(2x) + C\]
I hope this makes sense!
