Example 6. Find $\int \frac{d x}{2 x^{2}-x-1}$
Solution. The form of the denominator of the fraction (it has two real roots) suggests the use of formula 14. Therefore,

$$
\begin{aligned}
& \int \frac{d x}{2 x^{2}-x-1}=\int \frac{d x}{2\left(x^{2}-\frac{1}{2} x-\frac{1}{2}\right)}=\frac{1}{2} \int \frac{d x}{\left(x-\frac{1}{4}\right)^{2}-\frac{9}{16}}= \\
& \quad=\left\{\begin{array}{c}
u=x-\frac{1}{4} \\
m^{2}=\frac{9}{16} \\
m=\frac{1}{4}
\end{array}\right\}=\frac{1}{2} \cdot \frac{1}{2 \cdot \frac{3}{4}} \ln \left|\frac{\left.x-\frac{1}{4}-\frac{3}{4} \right\rvert\,}{x-\frac{1}{4}+\frac{3}{4}}\right|+C=\frac{1}{3} \ln \left|\frac{2(x-1)}{2 x+1}\right|+C
\end{aligned}
$$

Now let's apply the linearity property of indefinite integrals to tabular integration, which can be written as the equality

$$
\int(A f(x) \pm B g(x)) d x=A \int f(x) d x \pm B \int g(x) d x
$$