489. Find the integrals:
1) $\int \frac{d x}{\sqrt{x^{2}-4 x-3}}$
2) $\int \frac{(3 x-5) d x}{\sqrt{9+6 x-3 x^{2}}}$
Solution. 1) By completing the square of the trinomial $x^{2} - 4x - 3 = (x-2)^{2} - 7$, writing $d(x-2)$ instead of $dx$, and integrating, we find

$$
\int \frac{dx}{\sqrt{x^{2} - 4x - 3}} = \int \frac{d(x-2)}{\sqrt{(x-2)^{2} - 7}} = \ln \left| x-2 + \sqrt{(x-2)^{2} - 7} \right| + C
$$

(according to formula 11, with $u = x-2, a = -7$).

2) Completing the square of the quadratic trinomial $9 + 6x - 3x^{2} = -3(x^{2} - 2x - 3) = -3\left[(x-1)^{2} - 4\right] = 3\left[4 - (x-1)^{2}\right]$ and introducing a new temporary variable $z = x-1$. Then we get: $dx = dz$,

$$
I = \int \frac{(3x-5)dx}{\sqrt{9 + 6x - 3x^{2}}} = \frac{1}{\sqrt{3}} \int \frac{3z - 2}{\sqrt{4 - z^{2}}} dz
$$

By splitting the obtained integral into two integrals,

$$
I = \frac{3}{\sqrt{3}} \int \frac{z dz}{\sqrt{4 - z^{2}}} - \frac{2}{\sqrt{3}} \int \frac{dz}{\sqrt{4 - z^{2}}} = \sqrt{3} I_{1} - \frac{2}{\sqrt{3}} I_{2}
$$

we find each of them separately.

The first integral is transformed to formula 1 by multiplying and dividing it by -2 and replacing $-2z dz$ with $d(4 - z^{2})$:

$$
\begin{aligned}
& I_{1} = \int \frac{z dz}{\sqrt{4 - z^{2}}} = -\frac{1}{2} \int (4 - z^{2})^{-\frac{1}{2}} (-2z dz) = \\
& = -\frac{1}{2} \int (4 - z^{2})^{-\frac{1}{2}} d(4 - z^{2}) = - (4 - z^{2})^{\frac{1}{2}}
\end{aligned}
$$

The second integral is found using formula 10, with $u = z, a = 2$:

$$
I_{2} = \int \frac{dz}{\sqrt{4 - z^{2}}} = \arcsin \frac{z}{2}
$$

Substituting the found integrals $I_{1}$ and $I_{2}$ and returning to the variable $x$, we get

$$
\begin{aligned}
& I = C - \sqrt{3(4 - z^{2})} - \frac{2}{\sqrt{3}} \arcsin \frac{2}{2} = C - \sqrt{9 + 6x - 3x^{2}} - \\
& - \frac{2}{\sqrt{3}} \arcsin \frac{x-1}{2}
\end{aligned}
$$