Question

Интеграл (e^cos3x)*sin3xdx интеграл ln(x^2+5)dx интеграл x^3/(x^2+2x-1)dx

Answer 1

$\displaystyle \int (e^{cos3x}sin3xdx)=-\frac{1}{3}\int (e^{cos3x}d(cos3x)=-\frac{1}{3}e^{cos3x}+C$

$\displaystyle \int ln(x^2+5)dx=xln|x^2+5|-2\int\frac{x^2+5-5}{x^2+5}dx=xln|x^2+5|-\\-2\int(1-\frac{5}{x^2+5})dx=xln|x^2+5|-2x+2\sqrt5\ arctg\frac{x}{\sqrt5}+C\\\\\\\u=ln|x^2+5|;du=\frac{2x}{x^2+5}\\dv=dx;v=x$

$\displaystyle \int\frac{x^3dx}{x^2+2x-1}=\int\frac{(x^2+2x-1-2x+1)x}{x^2+2x-1}dx=\\=\int(x-\frac{2x^2+4x-2-5x+2}{x^2+2x-1})dx=\int(x-2+\frac{5x-2}{x^2+2x-1})dx\\\int(x-2+\frac{5}{2}\frac{2x+2-\frac{14}{5}}{x^2+2x-1})=\int(x-2)dx+\frac{5}{2}\int\frac{d(x^2+2x-1)}{x^2+2x-1}-\\-7\int\frac{d(x+1)}{(x+1)^2-2}=\frac{x^2}{2}-2x+\frac{5}{2}ln|x^2+2x-1|-\\-\frac{7}{2\sqrt2}ln|\frac{x+1-\sqrt2}{x+1+\sqrt2}|+C$