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

Ta có:

\(
f\left( 1 \right) = 0 \Rightarrow a + b = 0.\)

Do 

\(
f\left( x \right) = a{\mkern 1mu} x + \frac{b}{{{x^2}}}\left( {x \ne 0} \right) \Rightarrow F\left( x \right) = \mathop \smallint \limits_{}^{} f\left( x \right)dx = \frac{{a{\mkern 1mu} {x^2}}}{2} – \frac{b}{x} + C\)

Do

\(\begin{array}{*{20}{c}}
{}&{F\left( { – 1} \right) = 1 \Rightarrow \frac{a}{2} + b + C = 1}\\
{}&{F\left( 1 \right) = 4 \Rightarrow \frac{a}{2} – b + C = 4}
\end{array}\)

Suy ra 

\(
a = \frac{3}{2};b = – \frac{3}{2};c = \frac{7}{4} \Rightarrow F\left( x \right) = \frac{{3{x^2}}}{4} + \frac{3}{{2x}} + \frac{7}{4}\)