Question

If $af\left(x\right)+bf\left(\frac{1}{x}\right)=\frac{1}{x}-5,x\ne 0,a\ne b$, then ${\int }_1^{2}f\left(x\right)dx$ equals

Answer 1

$a+\left(x\right)+bf\left(\frac{1}{x}\right)=\frac{1}{x}-5\times a$

$x\to 1/x$

$af\left(\frac{1}{x}\right)+bf\left(x\right)=x-5\times b$

$a^{2}f\left(x\right)+abf\left(\frac{1}{x}\right)=\frac{a}{x}-5a$

$b^{2}f\left(x\right)+abf\left(\frac{1}{x}\right)=bx-5b$

_______________________

$(a^2-b^2)f\left(x\right)=\frac{a}{x}-bx-5a+5b$

$f\left(x\right)=\frac{\frac{a}{x}-bx-5a+5b}{a^2-b^2}$

${\int }_1^{2}f\left(x\right)dx={\int }_1^2\left(\frac{a/x-bx-5a+5b}{a^2-b^2}\right)dx$

$=\frac{[a\ln x-\frac{b}{2}{x}^2-5ax+5bx{]}_1^2}{a^2-b^2}$

$=\frac{a(ln2-5)+5b-\frac{3b}{2}}{a^2-b^2}$

$=\frac{a(ln2-5)+7b/2}{a^2-b^2}$