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4x-2-5-2x-2-4-dx-x-0-




Question Number 4594 by alimyao last updated on 10/Feb/16
∫4x^2  − (5/(2x^(−2) )) + 4  dx, x≠0
$$\int\mathrm{4}{x}^{\mathrm{2}} \:−\:\frac{\mathrm{5}}{\mathrm{2}{x}^{−\mathrm{2}} }\:+\:\mathrm{4}\:\:{dx},\:{x}\neq\mathrm{0} \\ $$$$ \\ $$
Answered by FilupSmith last updated on 10/Feb/16
=4∫x^2 dx−(5/2)∫x^2 dx+4∫dx  =(4/3)x^3 −(5/6)x^3 +4x+c  =(1/2)x^3 +4x+c
$$=\mathrm{4}\int{x}^{\mathrm{2}} {dx}−\frac{\mathrm{5}}{\mathrm{2}}\int{x}^{\mathrm{2}} {dx}+\mathrm{4}\int{dx} \\ $$$$=\frac{\mathrm{4}}{\mathrm{3}}{x}^{\mathrm{3}} −\frac{\mathrm{5}}{\mathrm{6}}{x}^{\mathrm{3}} +\mathrm{4}{x}+{c} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}{x}^{\mathrm{3}} +\mathrm{4}{x}+{c} \\ $$

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