Question

∫(2x-3)/sqrt(2-6x+9x^2)dx ∫(4x+1)/(6x^2-2x+3)dx ∫(x+1)/(2x^2-5x+2)dx

Answer 1

$\int \frac{2x-3}{\sqrt{2-6x+9x^2}}dx=\frac{1}{9}\int\frac{18x-6-21}{\sqrt{2-6x+9x^2}}dx=\frac{1}{9}\int\frac{d(2-6x+9x^2)}{\sqrt{2-6x+9x^2}}-\\-\frac{7}{9} \int\frac{d(3x-1)}{\sqrt{(3x-1)^2+1}}=\frac{2}{9}\sqrt{9x^2-6x+2}-\\-\frac{7}{9}ln|3x-1+\sqrt{(3x-1)^2+1}|+C\\\\\\(2-6x+9x^2)'=-6+18x$

$\int\frac{4x+1}{6x^2-2x+3}dx=\frac{1}{3}\int\frac{12x-2+5}{6x^2-2x+3}dx=\frac{1}{3}\int\frac{d(6x^2-2x+3)}{6x^2-2x+3}+\\+\frac{5}{3\sqrt6}\int\frac{d(\sqrt{6}x-\frac{1}{\sqrt6})}{(\sqrt{6}x-\frac{1}{\sqrt6})^2+\frac{17}{6}}=\frac{1}{3}ln|6x^2-2x+3|+\\+\frac{5}{3\sqrt{17}}arctg\frac{6x-1}{\sqrt{17}}+C\\\\\\(6x^2-2x+3)'=12x-2$

$\int\frac{x+1}{2x^2-5x+2}dx=\frac{1}{4}\int\frac{4x-5+9}{2x^2-5x+2}dx=\frac{1}{4}\int\frac{d(2x^2-5x+2)}{2x^2-5x+2}+\\+\frac{9}{4\sqrt2}\int\frac{d(\sqrt2x-\frac{5}{2\sqrt2})}{(\sqrt2x-\frac{5}{2\sqrt2})^2-\frac{9}{8}}=\frac{1}{4}ln|2x^2-5x+2|+\frac{3}{4}ln|\frac{2x-4}{2x-1}|+C\\\\\\(2x^2-5x+2)'=4x-5$