Question Number 26738 by goswamisubhabrata007@gmail.com last updated on 28/Dec/17
Commented by prakash jain last updated on 28/Dec/17
![∫e^x (log x+(1/x))dx ∫e^x [f(x)+f′(x)]dx =∫e^x f(x)dx+∫e^x f′(x)dx =f(x)e^x −∫e^x f′(x)dx+∫e^x f′(x)dx =f(x)e^x +C hence ∫e^x (log x+(1/x))dx=e^x log x+C](Q26747.png)
$$\int{e}^{{x}} \left(\mathrm{log}\:{x}+\frac{\mathrm{1}}{{x}}\right){dx} \\ $$$$\int{e}^{{x}} \left[{f}\left({x}\right)+{f}'\left({x}\right)\right]{dx} \\ $$$$=\int{e}^{{x}} {f}\left({x}\right){dx}+\int{e}^{{x}} {f}'\left({x}\right){dx} \\ $$$$={f}\left({x}\right){e}^{{x}} −\int{e}^{{x}} {f}'\left({x}\right){dx}+\int{e}^{{x}} {f}'\left({x}\right){dx} \\ $$$$={f}\left({x}\right){e}^{{x}} +{C} \\ $$$${hence} \\ $$$$\int{e}^{{x}} \left(\mathrm{log}\:{x}+\frac{\mathrm{1}}{{x}}\right){dx}={e}^{{x}} \mathrm{log}\:{x}+{C} \\ $$
Answered by Penguin last updated on 28/Dec/17
$$\int\frac{{e}^{{x}} }{{x}}\left({x}\mathrm{ln}\left({x}\right)+\mathrm{1}\right){dx}=\int{e}^{{x}} \mathrm{ln}\left({x}\right){dx}+\int\frac{{e}^{{x}} }{{x}}{dx} \\ $$$$\: \\ $$$$\mathrm{solve}\:\int{e}^{{x}} \mathrm{ln}\left({x}\right){dx}\:\mathrm{via}\:\mathrm{integration}\:\mathrm{by}\:\mathrm{parts} \\ $$$${u}={e}^{{x}} \:\:\:\:\:\:\:\:{v}=\frac{\mathrm{1}}{{x}} \\ $$$${u}'={e}^{{x}} \:\:\:\:\:\:\:{v}'=\mathrm{ln}\left({x}\right) \\ $$$$\int{e}^{{x}} \mathrm{ln}\left({x}\right){dx}=\int{uv}'={uv}−\int{u}'{v} \\ $$$$={e}^{{x}} \mathrm{ln}\left({x}\right)−\int\frac{{e}^{{x}} }{{x}}{dx} \\ $$$$\: \\ $$$$\therefore\int{e}^{{x}} \mathrm{ln}\left({x}\right){dx}+\int\frac{{e}^{{x}} }{{x}}{dx}={e}^{{x}} \mathrm{ln}\left({x}\right)+\int\frac{{e}^{{x}} }{{x}}{dx}−\int\frac{{e}^{{x}} }{{x}}{dx} \\ $$$$\therefore\int\frac{{e}^{{x}} }{{x}}\left({x}\mathrm{ln}\left({x}\right)+\mathrm{1}\right){dx}={e}^{{x}} \mathrm{ln}\left({x}\right)+{c} \\ $$