Question

1) ∫ dx/корень5x+2
2)∫ (5x^4+14x^2+9)/(x^2+2)dx
3)∫(3корня из x - 1/x^3 - x^8)dx
4)∫(3-2x)/(2+x)dx
5)∫(1+3x)/(корень 1-4x^2+2x)dx
6)∫ (arctg2x)/(1+4x^2)dx
7)∫(x^2dx)/(sin^2 5x^3)

Answer 1

$\displaystyle\int \frac{dx}{\sqrt{5x+2}}=\frac{1}{5}\int\frac{d(5x+2)}{\sqrt{5x+2}}=\frac{2}{5}\sqrt{5x+2}+C$

$\displaystyle\int \frac{5x^4+14x^2+9}{x^2+2}dx=\int(5x^2+4+\frac{1}{x^2+2})dx=\frac{5x^3}{3}+4x+\\+\frac{1}{\sqrt2}arctg\frac{x}{\sqrt2}+C$

$\displaystyle \int(3\sqrt{x}-\frac{1}{x^3}-x^8)dx=2\sqrt{x^3}+\frac{1}{2x^2}-\frac{x^9}{9}+C$

$\displaystyle \int\frac{3-2x}{2+x}dx=-2\int\frac{-\frac{7}{2}+x+2}{2+x}=7\int\frac{d(x+2)}{x+2}-2\int dx=\\=7ln|x+2|-2x+C$

$\displaystyle \int \frac{1+3x}{\sqrt{1-4x^2+2x}}dx=-\frac{3}{8}\int\frac{-\frac{14}{3}-8x+2}{\sqrt{1-4x^2+2x}}=\\=-\frac{3}{8}\int\frac{d(1-4x^2+2x)}{\sqrt{1-4x^2+2x}}+\frac{7}{8}\int\frac{d(2x-\frac{1}{2})}{\sqrt{-(2x-\frac{1}{2})^2+\frac{5}{4}}}=\\=-\frac{3}{4}\sqrt{1-4x^2+2x}+\frac{7}{8}arcsin\frac{\sqrt{4}(4x-1)}{2\sqrt{5}}+C\\(1-4x^2+2x)'=-8x+2$

$\displaystyle \int\frac{arctg2x}{1+4x^2}dx=\frac{1}{2}\int arctg2xd(arctg2x)=\frac{arctg^22x}{4}+C$

$\displaystyle \int \frac{x^2dx}{sin^25x^3}=\frac{1}{15}\int\frac{d(5x^3)}{sin^25x^3}=-\frac{1}{15}ctg5x^3+C$