Question

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

Answer 1

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

$\displaystyle\int\frac{dx}{\sqrt{4-9x^2}}=\frac{1}{3}\int\frac{d(\frac{3x}{2})}{\sqrt{1-(\frac{3x}{2})^2}}=\frac{1}{3}arcsin\frac{3x}{2}+C$

$\displaystyle\int\frac{dx}{5x-3}=\frac{1}{5}\int\frac{d(5x-3)}{5x-3}=\frac{1}{5}ln|5x-3|+C$