Find all the antiderivatives of \(f(x)=e^{x}+\dfrac{6}{x^{2}}-\sin x\).
Since the antiderivatives of \(e^{x}\) are \(e^{x}+{C}_{1}\), the antiderivatives of \(\tfrac{6}{x^{2}}\) are \(-\tfrac{6}{x}+{C}_{2}\), and the antiderivatives of \(\sin\) \(x\) are \(-\cos\) \(x + {C}_{3}\), the constant \(C\) in Example 2 is actually the sum of the constants \(C_{1},\) \(C_{2},\) and \(C_{3}.\)
An antiderivative of \(e^{x}\) is \(e^{x}.\) An antiderivative of \(6x^{-2}\) is \[ 6\cdot \frac{x^{-2+1}}{-2+1}=6\cdot \dfrac{x^{-1}}{-1}=-\dfrac{6}{x} \]
Finally, an antiderivative of \(\sin x\) is \(-\;\cos x\). Then all the antiderivatives of the function \(f\) are given by \[ F(x) =e^{x}-\frac{6}{x}+\cos x+C \]
where \(C\) is a constant.