Tìm nguyên hàm của hàm số f(x)=ax+b/x^2 biết F(-1)=1, F(1)=4, f(1)=0

$\begin{array}{l} \int {f(x)dx} = \int {\left( {ax + \frac{b}{{{x^2}}}} \right)dx} = \int {\left( {ax + b{x^{ - 2}}} \right)dx} \\ = \frac{{a{x^2}}}{2} + \frac{{b{x^{ - 1}}}}{{ - 1}} + C = \frac{{a{x^2}}}{2} - \frac{b}{x} + C = F(x). \end{array}$

Ta có: $\left\{ \begin{array}{l} F( - 1) = 1\\ F(1) = 4\\ f(1) = 0 \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} \frac{a}{2} + b + C = 1\\ \frac{a}{2} - b + C = 4\\ a + b = 0 \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} a = \frac{3}{2}\\ b = - \frac{3}{2}\\ c = \frac{7}{4} \end{array} \right.$  

Vậy: $F(x) = \frac{{3{x^2}}}{4} + \frac{3}{{2x}} + \frac{7}{4}$