Question

Найти интеграл :
7)xdx/16x^2+9
8)dx/√4-9x^2 (кв. корень над всем )
9)dx/5x-3

Answer 1

$displaystyle intfrac{xdx}{16x^2+9}=frac{1}{32}intfrac{d(16x^2+9)}{16x^2+9}=frac{1}{32}ln|16x^2+9|+C$

$displaystyleintfrac{dx}{sqrt{4-9x^2}}=frac{1}{3}intfrac{d(frac{3x}{2})}{sqrt{1-(frac{3x}{2})^2}}=frac{1}{3}arcsinfrac{3x}{2}+C$

$displaystyleintfrac{dx}{5x-3}=frac{1}{5}intfrac{d(5x-3)}{5x-3}=frac{1}{5}ln|5x-3|+C$