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

$displaystyleint frac{dx}{sqrt{5x+2}}=frac{1}{5}intfrac{d(5x+2)}{sqrt{5x+2}}=frac{2}{5}sqrt{5x+2}+C$

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

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

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

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

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

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