Question

Помогите найти первообразную, сроооочно

Answer 1

$\displaystyle\int(\frac{1}{(3-4x)^7}+\frac{2x}{\sqrt{6x-3}})dx=-\frac{1}{4}\int\frac{d(3-4x)}{(3-4x)^7}+\\+\frac{1}{\sqrt{3}}\int\frac{2x-1+1}{\sqrt{2x-1}}dx=-\frac{1}{4}\int\frac{d(3-4x)}{(3-4x)^7}+\\+\frac{1}{2\sqrt3}\int\sqrt{2x-1}d(2x-1)+\frac{1}{2\sqrt{3}}\int\frac{d(2x-1)}{\sqrt{2x-1}}=\\=\frac{1}{24(3-4x)^6}+\frac{\sqrt{(2x-1)^3}}{3\sqrt3}+\sqrt{\frac{2x-1}{3}}+C$

$\displaystyle\int (cos^4\frac{x}{3}+sin^6\frac{x}{3})dx\\\int cos^4\frac{x}{3}dx=\frac{1}{4}\int(\frac{3}{2}+2cos\frac{2x}{3}+\frac{1}{2}cos\frac{4x}{3})dx=\\=\frac{3}{8}\int dx+\frac{3}{4}\int cos\frac{2x}{3}d(\frac{2x}{3})+\frac{3}{32}\int cos\frac{4x}{3}d(\frac{4x}{3})=\\=\frac{3x}{8}+\frac{3}{4}sin\frac{2x}{3}+\frac{3}{32}sin\frac{4x}{3}+C$
$\displaystyle \int sin^6\frac{x}{3}dx=\frac{1}{8}\int(1-cos\frac{2x}{3})^3dx=\\=\frac{1}{8}\int(1-3cos\frac{2x}{3}+3cos^2\frac{2x}{3}-cos^3\frac{2x}{3})dx=\\\frac{1}{8}\int(\frac{5}{2}-4cos\frac{2x}{3}+\frac{3}{2}cos\frac{4x}{3}+sin^2\frac{2x}{3}cos\frac{2x}{3})dx=\\=\frac{5}{16}x-\frac{3}{4}sin\frac{2x}{3}+\frac{9}{64}sin\frac{4x}{3}+\frac{1}{16}sin^3\frac{2x}{3}+C$
$\displaystyle\int (cos^4\frac{x}{3}+sin^6\frac{x}{3})dx=\frac{11x}{16}+\frac{15}{64}sin\frac{4x}{3}+\frac{1}{16}sin^3\frac{2x}{3}+C$

$\displaystyle\int(\frac{6x^3}{\sqrt{6x+4x^2}}+\frac{1}{9}\sqrt{9x-2})dx\\\int\frac{6x^3}{\sqrt{6x+4x^2}}dx=6\int\frac{x^2\sqrt{6x+4x^2}}{4x+6}=\\=6\int(\frac{x}{4}-\frac{6}{16}+\frac{9}{4}\frac{1}{4x+6})\sqrt{6x+4x^2}dx=\\=\frac{6}{16}\int(4x-6)\sqrt{6x+4x^2}dx+\frac{27}{2}\int\frac{\sqrt{x}dx}{\sqrt{4x+6}}=\\=\frac{1}{8}\sqrt{(6x+4x^2)^3}-\frac{27}{32}((2x+\frac{3}{2})\sqrt{4x^2+6x}-\\-\frac{9}{4}ln|2x+\frac{3}{2}+\sqrt{4x^2+6x}|)+\frac{1}{8}(\sqrt{4x+6}\sqrt{4x}-\\-6ln|\sqrt{4x+6}+\sqrt{4x}|)+C$
$\int\frac{1}{9}\sqrt{9x-2}dx=\frac{1}{81}\int\sqrt{9x-2}d(9x+2)=\frac{2}{243}\sqrt{(9x-2)^3}+C$
$\displaystyle\int(\frac{6x^3}{\sqrt{6x+4x^2}}+\frac{1}{9}\sqrt{9x-2})dx=\frac{1}{8}\sqrt{(6x+4x^2)^3}-\\-\frac{27}{32}((2x+\frac{3}{2})\sqrt{4x^2+6x}-\frac{9}{4}ln|2x+\frac{3}{2}+\sqrt{4x^2+6x}|)+\\+\frac{1}{8}(\sqrt{4x+6}\sqrt{4x}-6ln|\sqrt{4x+6}+\sqrt{4x}|)+\frac{2}{243}\sqrt{(9x-2)^3}+C$