Question Number 74446 by rajesh4661kumar@gmail.com last updated on 24/Nov/19
Answered by Kunal12588 last updated on 24/Nov/19
![I=∫e^(tan^(−1) x) (((1+x+x^2 )/(1+x^2 )))dx let t=tan^(−1) x ⇒dt=(dx/(1+x^2 )) I=∫e^t (1+tan t + tan^2 t)dt ⇒I=∫e^t (tan t + sec^2 t)dt ⊛ ⇒I=e^t tan t +C ⇒I=xe^(tan^(−1) x) +C ⊛ ∫e^x [f(x)+f ′(x)]dx=e^x f(x)+C](https://www.tinkutara.com/question/Q74452.png)
$${I}=\int{e}^{{tan}^{−\mathrm{1}} {x}} \left(\frac{\mathrm{1}+{x}+{x}^{\mathrm{2}} }{\mathrm{1}+{x}^{\mathrm{2}} }\right){dx} \\ $$$${let}\:{t}={tan}^{−\mathrm{1}} {x} \\ $$$$\Rightarrow{dt}=\frac{{dx}}{\mathrm{1}+{x}^{\mathrm{2}} } \\ $$$${I}=\int{e}^{{t}} \left(\mathrm{1}+{tan}\:{t}\:+\:{tan}^{\mathrm{2}} \:{t}\right){dt} \\ $$$$\Rightarrow{I}=\int{e}^{{t}} \left({tan}\:{t}\:+\:{sec}^{\mathrm{2}} \:{t}\right){dt}\:\:\circledast \\ $$$$\Rightarrow{I}={e}^{{t}} {tan}\:{t}\:+{C} \\ $$$$\Rightarrow{I}={xe}^{{tan}^{−\mathrm{1}} {x}} \:+{C} \\ $$$$\circledast\:\int{e}^{{x}} \left[{f}\left({x}\right)+{f}\:'\left({x}\right)\right]{dx}={e}^{{x}} {f}\left({x}\right)+{C} \\ $$
Commented by Kunal12588 last updated on 24/Nov/19
![∫e^x [f(x)+f ′(x)]dx =∫e^x f(x)dx+∫e^x f ′(x) dx =e^x f(x)−∫e^x f ′(x)dx+∫e^x f ′(x)dx =e^x f(x)+C](https://www.tinkutara.com/question/Q74454.png)
$$\int{e}^{{x}} \left[{f}\left({x}\right)+{f}\:'\left({x}\right)\right]{dx} \\ $$$$=\int{e}^{{x}} {f}\left({x}\right){dx}+\int{e}^{{x}} {f}\:'\left({x}\right)\:{dx} \\ $$$$={e}^{{x}} {f}\left({x}\right)−\int{e}^{{x}} {f}\:'\left({x}\right){dx}+\int{e}^{{x}} {f}\:'\left({x}\right){dx} \\ $$$$={e}^{{x}} {f}\left({x}\right)+{C} \\ $$