Question Number 144026 by mohammad17 last updated on 20/Jun/21
Commented by mohammad17 last updated on 20/Jun/21
$${help}\:{me}\:{sir} \\ $$
Answered by Olaf_Thorendsen last updated on 20/Jun/21
![Ω = ∫_0 ^(ln2) ((e^(5x) +2e^(4x) +e^(2x) +e^x )/(e^(2x) +e^x +1)) dx Let u = e^x Ω = ∫_1 ^2 ((u^5 +2u^4 +u^2 +u)/(u^2 +u+1)). (du/u) Ω = ∫_1 ^2 ((u^4 +2u^3 +u+1)/(u^2 +u+1)) du Ω = ∫_1 ^2 (((u^2 +u−2)(u^2 +u+1)+(2u+1)+2)/(u^2 +u+1)) du Ω = ∫_1 ^2 (u^2 +u−2+((2u+1)/(u^2 +u+1))+(2/(u^2 +u+1))) du Ω_1 = ∫_1 ^2 (u^2 +u−2) du Ω_1 = [(u^3 /3)+(u^2 /2)−2u]_1 ^2 Ω_1 = ((8/3)+2−4)−((1/3)+(1/2)−2) Ω_1 = (4/6)−(−(7/6)) = ((11)/6) Ω_2 = ∫_1 ^2 ((2u+1)/(u^2 +u+1)) du Ω_2 = [ln∣u^2 +u+1∣]_1 ^2 Ω_2 = ln7−ln3 = ln(7/3) Ω_3 = ∫_1 ^2 (2/(u^2 +u+1)) du Ω_3 = ∫_1 ^2 (2/((u+(1/2))^2 +(3/4))) du Ω_3 = 2[(1/((√3)/2))arctan(((u+(1/2))/((√3)/2)))]_1 ^2 Ω_3 = (4/( (√3)))[arctan((2/( (√3)))(u+(1/2))]_1 ^2 Ω_3 = (4/( (√3)))(arctan((5/( (√3))))−arctan((3/( (√3))))) Ω_3 = (4/( (√3)))arctan((((5/( (√3)))−(3/( (√3))))/(1+(5/( (√3))).(3/( (√3)))))) Ω_3 = (4/( (√3)))arctan(((2/( (√3)))/6)) Ω_3 = (4/( (√3)))arctan((1/(3(√3)))) Ω =Ω_1 +Ω_2 +Ω_3 Ω = ((11)/6)+ln(7/3)+(4/( (√3)))arctan((1/(3(√3))))](https://www.tinkutara.com/question/Q144032.png)
$$\Omega\:=\:\int_{\mathrm{0}} ^{\mathrm{ln2}} \frac{{e}^{\mathrm{5}{x}} +\mathrm{2}{e}^{\mathrm{4}{x}} +{e}^{\mathrm{2}{x}} +{e}^{{x}} }{{e}^{\mathrm{2}{x}} +{e}^{{x}} +\mathrm{1}}\:{dx} \\ $$$$\mathrm{Let}\:{u}\:=\:{e}^{{x}} \\ $$$$\Omega\:=\:\int_{\mathrm{1}} ^{\mathrm{2}} \frac{{u}^{\mathrm{5}} +\mathrm{2}{u}^{\mathrm{4}} +{u}^{\mathrm{2}} +{u}}{{u}^{\mathrm{2}} +{u}+\mathrm{1}}.\:\frac{{du}}{{u}} \\ $$$$\Omega\:=\:\int_{\mathrm{1}} ^{\mathrm{2}} \frac{{u}^{\mathrm{4}} +\mathrm{2}{u}^{\mathrm{3}} +{u}+\mathrm{1}}{{u}^{\mathrm{2}} +{u}+\mathrm{1}}\:{du} \\ $$$$\Omega\:=\:\int_{\mathrm{1}} ^{\mathrm{2}} \frac{\left({u}^{\mathrm{2}} +{u}−\mathrm{2}\right)\left({u}^{\mathrm{2}} +{u}+\mathrm{1}\right)+\left(\mathrm{2}{u}+\mathrm{1}\right)+\mathrm{2}}{{u}^{\mathrm{2}} +{u}+\mathrm{1}}\:{du} \\ $$$$\Omega\:=\:\int_{\mathrm{1}} ^{\mathrm{2}} \left({u}^{\mathrm{2}} +{u}−\mathrm{2}+\frac{\mathrm{2}{u}+\mathrm{1}}{{u}^{\mathrm{2}} +{u}+\mathrm{1}}+\frac{\mathrm{2}}{{u}^{\mathrm{2}} +{u}+\mathrm{1}}\right)\:{du} \\ $$$$\Omega_{\mathrm{1}} \:=\:\int_{\mathrm{1}} ^{\mathrm{2}} \left({u}^{\mathrm{2}} +{u}−\mathrm{2}\right)\:{du} \\ $$$$\Omega_{\mathrm{1}} \:=\:\left[\frac{{u}^{\mathrm{3}} }{\mathrm{3}}+\frac{{u}^{\mathrm{2}} }{\mathrm{2}}−\mathrm{2}{u}\right]_{\mathrm{1}} ^{\mathrm{2}} \\ $$$$\Omega_{\mathrm{1}} \:=\:\left(\frac{\mathrm{8}}{\mathrm{3}}+\mathrm{2}−\mathrm{4}\right)−\left(\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{2}}−\mathrm{2}\right) \\ $$$$\Omega_{\mathrm{1}} \:=\:\frac{\mathrm{4}}{\mathrm{6}}−\left(−\frac{\mathrm{7}}{\mathrm{6}}\right)\:=\:\frac{\mathrm{11}}{\mathrm{6}} \\ $$$$\Omega_{\mathrm{2}} \:=\:\int_{\mathrm{1}} ^{\mathrm{2}} \frac{\mathrm{2}{u}+\mathrm{1}}{{u}^{\mathrm{2}} +{u}+\mathrm{1}}\:{du} \\ $$$$\Omega_{\mathrm{2}} \:=\:\left[\mathrm{ln}\mid{u}^{\mathrm{2}} +{u}+\mathrm{1}\mid\right]_{\mathrm{1}} ^{\mathrm{2}} \\ $$$$\Omega_{\mathrm{2}} \:=\:\mathrm{ln7}−\mathrm{ln3}\:=\:\mathrm{ln}\frac{\mathrm{7}}{\mathrm{3}} \\ $$$$ \\ $$$$\Omega_{\mathrm{3}} \:=\:\int_{\mathrm{1}} ^{\mathrm{2}} \frac{\mathrm{2}}{{u}^{\mathrm{2}} +{u}+\mathrm{1}}\:{du} \\ $$$$\Omega_{\mathrm{3}} \:=\:\int_{\mathrm{1}} ^{\mathrm{2}} \frac{\mathrm{2}}{\left({u}+\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} +\frac{\mathrm{3}}{\mathrm{4}}}\:{du} \\ $$$$\Omega_{\mathrm{3}} \:=\:\mathrm{2}\left[\frac{\mathrm{1}}{\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}}\mathrm{arctan}\left(\frac{{u}+\frac{\mathrm{1}}{\mathrm{2}}}{\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}}\right)\right]_{\mathrm{1}} ^{\mathrm{2}} \\ $$$$\Omega_{\mathrm{3}} \:=\:\frac{\mathrm{4}}{\:\sqrt{\mathrm{3}}}\left[\mathrm{arctan}\left(\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}}\left({u}+\frac{\mathrm{1}}{\mathrm{2}}\right)\right]_{\mathrm{1}} ^{\mathrm{2}} \right. \\ $$$$\Omega_{\mathrm{3}} \:=\:\frac{\mathrm{4}}{\:\sqrt{\mathrm{3}}}\left(\mathrm{arctan}\left(\frac{\mathrm{5}}{\:\sqrt{\mathrm{3}}}\right)−\mathrm{arctan}\left(\frac{\mathrm{3}}{\:\sqrt{\mathrm{3}}}\right)\right) \\ $$$$\Omega_{\mathrm{3}} \:=\:\frac{\mathrm{4}}{\:\sqrt{\mathrm{3}}}\mathrm{arctan}\left(\frac{\frac{\mathrm{5}}{\:\sqrt{\mathrm{3}}}−\frac{\mathrm{3}}{\:\sqrt{\mathrm{3}}}}{\mathrm{1}+\frac{\mathrm{5}}{\:\sqrt{\mathrm{3}}}.\frac{\mathrm{3}}{\:\sqrt{\mathrm{3}}}}\right) \\ $$$$\Omega_{\mathrm{3}} \:=\:\frac{\mathrm{4}}{\:\sqrt{\mathrm{3}}}\mathrm{arctan}\left(\frac{\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}}}{\mathrm{6}}\right) \\ $$$$\Omega_{\mathrm{3}} \:=\:\frac{\mathrm{4}}{\:\sqrt{\mathrm{3}}}\mathrm{arctan}\left(\frac{\mathrm{1}}{\mathrm{3}\sqrt{\mathrm{3}}}\right) \\ $$$$ \\ $$$$\Omega\:=\Omega_{\mathrm{1}} +\Omega_{\mathrm{2}} +\Omega_{\mathrm{3}} \\ $$$$\Omega\:=\:\frac{\mathrm{11}}{\mathrm{6}}+\mathrm{ln}\frac{\mathrm{7}}{\mathrm{3}}+\frac{\mathrm{4}}{\:\sqrt{\mathrm{3}}}\mathrm{arctan}\left(\frac{\mathrm{1}}{\mathrm{3}\sqrt{\mathrm{3}}}\right) \\ $$
Commented by BHOOPENDRA last updated on 20/Jun/21
$${Nice}\:{solution}\:{sir} \\ $$