Infi-2 2

\begin{matrix} \int \frac{x^{3}}{x^{2} - 3 x + 3} d x = \int (x + 3) d x + 3 \int \frac{2 x - 3}{x^{2} - 3 x + 3} d x = \\ = \frac{x^{2}}{2} + 3 x + 3 \ln | x^{2} - 3 x + 3 | + C \end{matrix}

\begin{matrix} \int \frac{x}{x^{2} - 4 x + 8} d x = \frac{1}{2} \int \frac{2 x - 4}{x^{2} - 4 x + 8} d x + \frac{1}{2} \int \frac{4}{x^{2} - 4 x + 8} d x = \\ = \frac{1}{2} \ln | x^{2} - 4 x + 8 | + 2 \int \frac{1}{(x - 2)^{2} + 2^{2}} d x = \\ = \frac{1}{2} \ln | x^{2} - 4 x + 8 | + \arctan (\frac{x - 2}{2}) + C \end{matrix}

\begin{matrix} \int \frac{x + 3}{x^{2} - 3 x - 40} d x = \int \frac{x + 3}{(x + 5) (x - 8)} d x = \\ = \int \frac{A}{x + 5} d x + \int \frac{B}{x - 8} d x = \\ {\begin{matrix} A + B = 1 \\ - 8 A + 5 B = 3 \end{matrix} ⟹ {\begin{matrix} A = \frac{2}{13} \\ B = \frac{11}{13} \end{matrix} \\ ⟹ \int \frac{x + 3}{x^{2} - 3 x - 40} d x = \frac{2}{13} \ln | x + 5 | + \frac{11}{13} \ln | x - 8 | + C \end{matrix}

\begin{matrix} \int \frac{x^{3} - 2}{x^{4} - x} d x = \int \frac{x^{3} - 2}{x (x^{3} - 1)} d x = \int \frac{x^{3} - 2}{x (x - 1) (x^{2} + x + 1)} d x \\ \frac{1}{x (x - 1) (x^{2} + x + 1)} = \frac{A}{x} + \frac{B}{x - 1} + \frac{C x + D}{x^{2} + x + 1} \\ x^{3} - 2 = A (x^{3} - 1) + B (x^{3} + x^{2} + x) + C x^{3} - C x^{2} + D x^{2} - D x \\ ⟹ {\begin{matrix} A + B + C = 1 \\ B - C + D = 0 \\ B - D = 0 \\ A = 2 \end{matrix} ⟹ {\begin{matrix} A = 2 \\ B = - \frac{1}{3} \\ C = - \frac{2}{3} \\ D = - \frac{1}{3} \end{matrix} \\ ⟹ \int \frac{x^{3} - 2}{x^{4} - x} d x = 2 \int \frac{1}{x} d x - \frac{1}{3} \int \frac{1}{x - 1} d x - \frac{1}{3} \int \frac{2 x + 1}{x^{2} + x + 1} d x = \\ = 2 \ln | x | - \frac{1}{3} \ln | x - 1 | - \frac{1}{3} \ln | x^{2} + x + 1 | + C \end{matrix}

\begin{matrix} \int \frac{x^{6} + x + 1}{x^{4} + 5 x^{2} + 4} d x = \dots = \int x^{2} - 5 + \frac{21 x^{2} + x + 21}{x^{4} + 5 x^{2} + 4} d x = \\ = \int (x^{2} - 5) d x + \int \frac{21 x^{2} + x + 21}{(x^{2} + 1) (x^{2} + 4)} d x = \dots \end{matrix}

\begin{matrix} \begin{array}{cc} \sqrt{a^{2} - x^{2}} & x = a \cdot \sin t \\ \sqrt{a^{2} + x^{2}} & x = a \cdot \tan t \\ \sqrt{x^{2} - a^{2}} & x = \frac{a}{\cos^{2} t} \end{array} \\ For example: \\ \int \frac{1}{x^{2} \sqrt{4 - x^{2}}} d x \\ x = 2 \sin t ⟹ d x = 2 \cos t d t \\ \int \frac{1}{x^{2} \sqrt{4 - x^{2}}} d x = \int \frac{2 \cos t}{4 \sin^{2} t \sqrt{4 \cos^{2} t}} d t = \frac{1}{4} \int \frac{1}{\sin^{2} t} d t = - \frac{1}{4} \cot t + C \\ \sin^{2} t = \frac{x^{2}}{4} = 1 - \cos^{2} t ⟹ \cos t = \frac{\sqrt{1 - x^{2}}}{2} \\ ⟹ \int \frac{1}{x^{2} \sqrt{4 - x^{2}}} d x = \frac{- \sqrt{1 - x^{2}}}{4 x} + C \end{matrix}

\begin{matrix} \int \frac{1}{\sin x} d x = {\begin{matrix} t = \tan \frac{x}{2} \\ d x = \frac{2}{1 + t^{2}} d t \\ \sin x = \frac{2 t}{1 + t^{2}} \end{matrix}} = \int \frac{1 + t^{2}}{2 t} \frac{2}{1 + t^{2}} d t = \int \frac{1}{t} d t = \\ = \ln | \tan \frac{x}{2} | + C \\ \int \frac{1}{\cos x} d x = {\begin{matrix} t = \tan \frac{x}{2} \\ d x = \frac{2}{1 + t^{2}} d t \\ \cos x = \frac{1 - t^{2}}{1 + t^{2}} \end{matrix}} = \int \frac{1 + t^{2}}{1 - t^{2}} \frac{2}{1 + t^{2}} d t = \int \frac{2}{1 - t^{2}} d t = \\ = \ln | 1 - t | + \ln | 1 + t | + C \end{matrix}