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प्रश्न
Evaluate `int (1 + "x" + "x"^2/(2!))`dx
उत्तर
`int (1 + "x" + "x"^2/(2!))`dx
`= int 1 "dx" + int "x" "dx" + 1/(2!) int "x"^2 "dx"`
`= "x" + "x"^2/2 + 1/(2!) xx "x"^3/3 + "c"`
∴ `"x" + "x"^2/2 + "x"^3/6 + "c"`
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