Question

Решение интеграла ( подробно, пожалуйста)

Answer 1

$\displaystyle \int\frac{dx}{\sqrt{3x^2-3x+8}}=\int\frac{dx}{\sqrt{3(x-\frac{1}{2})^2+\frac{29}{4}}}=\frac{1}{\sqrt{3}}\int\frac{d(x-\frac{1}{2})}{\sqrt{(x-\frac{1}{2})^2+(\sqrt{\frac{29}{12}})^2}}=\\ \\ \\ =\frac{1}{\sqrt{3}}\ln\bigg|x-\frac{1}{2}+\sqrt{\left(x-\frac{1}{2}\right)^2+\frac{29}{12}}\bigg|+C$

В конце использовали длинный логарифм.

Answer 2

$\displaystyle \int \dfrac{dx}{\sqrt{3x^{2} - 3x + 8}}$

$\displaystyle \int \dfrac{dx}{\sqrt{3(x^{2} - x) + 8}}$

$\displaystyle \int \dfrac{dx}{\sqrt{3\left (x^{2} - x + \dfrac{1}{4} - \dfrac{1}{4} \right) + 8}}$

$\displaystyle \int \dfrac{dx}{\sqrt{3\left [\left(x - \dfrac{1}{2} \right)^{2} - \dfrac{1}{4} \right] + 8}}$

$\displaystyle \int \dfrac{dx}{\sqrt{3\left(x - \dfrac{1}{2} \right)^{2} - \dfrac{3}{4} + 8}}$

$\displaystyle \int \dfrac{dx}{\sqrt{3\left(x - \dfrac{1}{2} \right)^{2} + \dfrac{29}{4}}}$

$\displaystyle \int \dfrac{dx}{\sqrt{3\left[ \left(x - \dfrac{1}{2} \right)^{2} + \dfrac{29}{12}\right]}}$

$\displaystyle \int \dfrac{dx}{\sqrt{3}\sqrt{ \left(x - \dfrac{1}{2} \right)^{2} + \dfrac{29}{12}}}$

$\dfrac{1}{\sqrt{3}} \displaystyle \int \dfrac{dx}{\sqrt{ \left(x - \dfrac{1}{2} \right)^{2} + \dfrac{29}{12}}}$

$\dfrac{1}{\sqrt{3}} \ln \left |x- \dfrac{1}{2} + \sqrt \left(x - \dfrac{1}{2} \right)^{2} + \dfrac{29}{12}}} \right | + C$