Question-91302

Question Number 91302 by 174 last updated on 29/Apr/20

Answered by MJS last updated on 29/Apr/20

∫e^x (((x−1)^2 )/((x^2 +1)^2 ))dx= =∫(e^x /(x^2 +1))dx−2∫((e^x x)/((x^2 +1)^2 ))dx the 2^(nd) one by parts: −2∫((e^x x)/((x^2 +1)^2 ))dx= u′=(x/((x^2 +1)^2 )) → u=−(1/(2(x^2 +1))) v=v′=e^x =(e^x /(x^2 +1))−∫(e^x /(x^2 +1))dx so we have ∫(e^x /(x^2 +1))dx+(e^x /(x^2 +1))−∫(e^x /(x^2 +1))dx= =(e^x /(x^2 +1))+C

$$\int\mathrm{e}^{{x}} \frac{\left({x}−\mathrm{1}\right)^{\mathrm{2}} }{\left({x}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} }{dx}= \\ $$$$=\int\frac{\mathrm{e}^{{x}} }{{x}^{\mathrm{2}} +\mathrm{1}}{dx}−\mathrm{2}\int\frac{\mathrm{e}^{{x}} {x}}{\left({x}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} }{dx} \\ $$$$\mathrm{the}\:\mathrm{2}^{\mathrm{nd}} \:\mathrm{one}\:\mathrm{by}\:\mathrm{parts}: \\ $$$$−\mathrm{2}\int\frac{\mathrm{e}^{{x}} {x}}{\left({x}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} }{dx}= \\ $$$$\:\:\:\:\:{u}'=\frac{{x}}{\left({x}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} }\:\rightarrow\:{u}=−\frac{\mathrm{1}}{\mathrm{2}\left({x}^{\mathrm{2}} +\mathrm{1}\right)} \\ $$$$\:\:\:\:\:{v}={v}'=\mathrm{e}^{{x}} \\ $$$$=\frac{\mathrm{e}^{{x}} }{{x}^{\mathrm{2}} +\mathrm{1}}−\int\frac{\mathrm{e}^{{x}} }{{x}^{\mathrm{2}} +\mathrm{1}}{dx} \\ $$$$\mathrm{so}\:\mathrm{we}\:\mathrm{have} \\ $$$$\int\frac{\mathrm{e}^{{x}} }{{x}^{\mathrm{2}} +\mathrm{1}}{dx}+\frac{\mathrm{e}^{{x}} }{{x}^{\mathrm{2}} +\mathrm{1}}−\int\frac{\mathrm{e}^{{x}} }{{x}^{\mathrm{2}} +\mathrm{1}}{dx}= \\ $$$$=\frac{\mathrm{e}^{{x}} }{{x}^{\mathrm{2}} +\mathrm{1}}+\mathrm{C} \\ $$

Answered by $@ty@m123 last updated on 29/Apr/20

Formula: ∫e^x {f(x)+f ′(x)}dx=e^x f(x)+C Here, f(x)=(1/(1+x^2 ))

$${Formula}: \\ $$$$\int{e}^{{x}} \left\{{f}\left({x}\right)+{f}\:'\left({x}\right)\right\}{dx}={e}^{{x}} {f}\left({x}\right)+\mathrm{C} \\ $$$${Here}, \\ $$$${f}\left({x}\right)=\frac{\mathrm{1}}{\mathrm{1}+{x}^{\mathrm{2}} } \\ $$

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