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Question Number 144026 by mohammad17 last updated on 20/Jun/21

Commented by mohammad17 last updated on 20/Jun/21

$$helpmesir$$

Answered by Olaf_Thorendsen last updated on 20/Jun/21

$$\mathrm{\Omega }={\int }_0^{\ln 2}\frac{e^{5x}+2e^{4x}+e^{2x}+e^x}{e^{2x}+e^x+1}dx$$$$\mathrm{Let}u=e^x$$$$\mathrm{\Omega }={\int }_1^2\frac{u^5+2u^4+u^2+u}{u^2+u+1}.\frac{du}{u}$$$$\mathrm{\Omega }={\int }_1^2\frac{u^4+2u^3+u+1}{u^2+u+1}du$$$$\mathrm{\Omega }={\int }_1^2\frac{(u^2+u-2)(u^2+u+1)+(2u+1)+2}{u^2+u+1}du$$$$\mathrm{\Omega }={\int }_1^2(u^2+u-2+\frac{2u+1}{u^2+u+1}+\frac{2}{u^2+u+1})du$$$${\mathrm{\Omega }}_1={\int }_1^2(u^2+u-2)du$$$${\mathrm{\Omega }}_1={[\frac{u^3}{3}+\frac{u^2}{2}-2u]}_1^2$$$${\mathrm{\Omega }}_1=(\frac{8}{3}+2-4)-(\frac{1}{3}+\frac{1}{2}-2)$$$${\mathrm{\Omega }}_1=\frac{4}{6}-(-\frac{7}{6})=\frac{11}{6}$$$${\mathrm{\Omega }}_2={\int }_1^2\frac{2u+1}{u^2+u+1}du$$$${\mathrm{\Omega }}_2={[\ln \mid u^2+u+1\mid ]}_1^2$$$${\mathrm{\Omega }}_2=\ln 7-\ln 3=\ln \frac{7}{3}$$$$$$$${\mathrm{\Omega }}_3={\int }_1^2\frac{2}{u^2+u+1}du$$$${\mathrm{\Omega }}_3={\int }_1^2\frac{2}{{(u+\frac{1}{2})}^2+\frac{3}{4}}du$$$${\mathrm{\Omega }}_3=2{\left[\frac{1}{\frac{\sqrt{3}}{2}}\arctan \left(\frac{u+\frac{1}{2}}{\frac{\sqrt{3}}{2}}\right)\right]}_1^2$$$${\mathrm{\Omega }}_3=\frac{4}{\sqrt{3}}[\arctan {\left(\frac{2}{\sqrt{3}}(u+\frac{1}{2})\right]}_1^2$$$${\mathrm{\Omega }}_3=\frac{4}{\sqrt{3}}(\arctan \left(\frac{5}{\sqrt{3}}\right)-\arctan \left(\frac{3}{\sqrt{3}}\right))$$$${\mathrm{\Omega }}_3=\frac{4}{\sqrt{3}}\arctan \left(\frac{\frac{5}{\sqrt{3}}-\frac{3}{\sqrt{3}}}{1+\frac{5}{\sqrt{3}}.\frac{3}{\sqrt{3}}}\right)$$$${\mathrm{\Omega }}_3=\frac{4}{\sqrt{3}}\arctan \left(\frac{\frac{2}{\sqrt{3}}}{6}\right)$$$${\mathrm{\Omega }}_3=\frac{4}{\sqrt{3}}\arctan \left(\frac{1}{3\sqrt{3}}\right)$$$$$$$$\mathrm{\Omega }={\mathrm{\Omega }}_1+{\mathrm{\Omega }}_2+{\mathrm{\Omega }}_3$$$$\mathrm{\Omega }=\frac{11}{6}+\ln \frac{7}{3}+\frac{4}{\sqrt{3}}\arctan \left(\frac{1}{3\sqrt{3}}\right)$$

Commented by BHOOPENDRA last updated on 20/Jun/21

$$Nicesolutionsir$$

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