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Question Number 26909 by sorour87 last updated on 31/Dec/17

∫((e^x ((1/x)−(2/x^3 )))/(−2))dx

$$\int\frac{{e}^{{x}} \left(\frac{\mathrm{1}}{{x}}−\frac{\mathrm{2}}{{x}^{\mathrm{3}} }\right)}{−\mathrm{2}}{dx} \\ $$

Answered by prakash jain last updated on 31/Dec/17

∫e^x (−(1/(2x))+(1/x^3 ))dx  ∫−(e^x /(2x))dx+∫(e^x /x^3 )dx  integrate first by parts  =−(e^x /(2x))−∫(e^x /(2x^2 ))dx+∫(e^x /x^3 )dx  =−(e^x /(2x))−(e^x /(2x^2 ))−∫(e^x /x^3 )dx+∫(e^x /x^3 )dx  =−((e^x (x+1))/(2x^2 ))

$$\int{e}^{{x}} \left(−\frac{\mathrm{1}}{\mathrm{2}{x}}+\frac{\mathrm{1}}{{x}^{\mathrm{3}} }\right){dx} \\ $$$$\int−\frac{{e}^{{x}} }{\mathrm{2}{x}}{dx}+\int\frac{{e}^{{x}} }{{x}^{\mathrm{3}} }{dx} \\ $$$${integrate}\:{first}\:{by}\:{parts} \\ $$$$=−\frac{{e}^{{x}} }{\mathrm{2}{x}}−\int\frac{{e}^{{x}} }{\mathrm{2}{x}^{\mathrm{2}} }{dx}+\int\frac{{e}^{{x}} }{{x}^{\mathrm{3}} }{dx} \\ $$$$=−\frac{{e}^{{x}} }{\mathrm{2}{x}}−\frac{{e}^{{x}} }{\mathrm{2}{x}^{\mathrm{2}} }−\int\frac{{e}^{{x}} }{{x}^{\mathrm{3}} }{dx}+\int\frac{{e}^{{x}} }{{x}^{\mathrm{3}} }{dx} \\ $$$$=−\frac{{e}^{{x}} \left({x}+\mathrm{1}\right)}{\mathrm{2}{x}^{\mathrm{2}} } \\ $$

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