Tìm một nguyên hàm \(F\left( x \right)\) của hàm số \(f\left( x \right) = ax + \frac{b}{{{x^2}}}\left( {x \ne 0} \right)\), biết \(F\left( { - 1} \right) = 1\), \(F\left( 1 \right) = 4,f\left( 1 \right) = 0\).

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

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

Vậy \(F\left( x \right) = \frac{{3{x^2}}}{4} + \frac{3}{{2x}} + \frac{7}{4}\).