Question Number 137940 by benjo_mathlover last updated on 08/Apr/21
$$\int_{\mathrm{1}} ^{\:{e}} \sqrt{\mathrm{1}+\frac{\mathrm{1}}{{x}^{\mathrm{2}} }+\frac{\mathrm{1}}{\left({x}+\mathrm{1}\right)^{\mathrm{2}\:} }}\:{dx}\:=? \\ $$
Answered by john_santu last updated on 08/Apr/21
![J = ∫_1 ^( e) (√(((x^2 +1)/x^2 )+(1/((x+1)^2 )))) dx J= ∫_1 ^( e) (√(((x^2 +1)(x+1)^2 +x^2 )/(x^2 (x+1)^2 ))) dx J= ∫_1 ^( e) (√(((x^2 +x+1)^2 )/(x^2 (x+1)^2 ))) dx J=∫_1 ^( e) ((x^2 +x+1)/(x(x+1))) dx J= ∫_1 ^( e) (1+(1/x)−(1/(x+1)))dx = [ x +ln ∣x∣−ln ∣x+1∣ ]_1 ^e = (e+1−ln (e+1))−(1−ln 2) =e+ln ((2/(e+1)))](https://www.tinkutara.com/question/Q137942.png)
$$\mathcal{J}\:=\:\int_{\mathrm{1}} ^{\:{e}} \sqrt{\frac{{x}^{\mathrm{2}} +\mathrm{1}}{{x}^{\mathrm{2}} }+\frac{\mathrm{1}}{\left({x}+\mathrm{1}\right)^{\mathrm{2}} }}\:{dx} \\ $$$$\mathcal{J}=\:\int_{\mathrm{1}} ^{\:{e}} \sqrt{\frac{\left({x}^{\mathrm{2}} +\mathrm{1}\right)\left({x}+\mathrm{1}\right)^{\mathrm{2}} +{x}^{\mathrm{2}} }{{x}^{\mathrm{2}} \left({x}+\mathrm{1}\right)^{\mathrm{2}} }}\:{dx} \\ $$$$\mathcal{J}=\:\int_{\mathrm{1}} ^{\:{e}} \:\sqrt{\frac{\left({x}^{\mathrm{2}} +{x}+\mathrm{1}\right)^{\mathrm{2}} }{{x}^{\mathrm{2}} \left({x}+\mathrm{1}\right)^{\mathrm{2}} }}\:{dx} \\ $$$$\mathcal{J}=\int_{\mathrm{1}} ^{\:{e}} \:\frac{{x}^{\mathrm{2}} +{x}+\mathrm{1}}{{x}\left({x}+\mathrm{1}\right)}\:{dx} \\ $$$$\mathcal{J}=\:\int_{\mathrm{1}} ^{\:{e}} \left(\mathrm{1}+\frac{\mathrm{1}}{{x}}−\frac{\mathrm{1}}{{x}+\mathrm{1}}\right){dx} \\ $$$$=\:\left[\:{x}\:+\mathrm{ln}\:\mid{x}\mid−\mathrm{ln}\:\mid{x}+\mathrm{1}\mid\:\right]_{\mathrm{1}} ^{{e}} \\ $$$$=\:\left({e}+\mathrm{1}−\mathrm{ln}\:\left({e}+\mathrm{1}\right)\right)−\left(\mathrm{1}−\mathrm{ln}\:\mathrm{2}\right) \\ $$$$={e}+\mathrm{ln}\:\left(\frac{\mathrm{2}}{{e}+\mathrm{1}}\right)\: \\ $$