Question

how do you solve ∫ [(3-x)/x(x^2+1)]. take note that (x^2)+1

Answer 1

\[\frac{3-x}{x(x^2+1)}=\frac{a}{x}+\frac{bx+c}{x^2+1}\]\[3-x=a(x^2+1)+x(bx+c)=(a+b)x^2+cx+a\]\[a=3\quad,\quad b=-3\quad,\quad c=-1\]\[\frac{3-x}{x(x^2+1)}=\frac{3}{x}-\frac{3x+1}{x^2+1}=\frac{3}{x}-\frac{3x}{x^2+1}-\frac{1}{x^2+1}\]\[\int\frac{3-x}{x(x^2+1)}dx=\int\left(\frac{3}{x}-\frac{3x}{x^2+1}-\frac{1}{x^2+1}\right)dx=\]\[=3\ln{x}-\frac{3}{2}\ln{(x^2+1)}-\arctan{x}+C\]