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Question Number 99077 by Mr.D.N. last updated on 18/Jun/20

Commented by MJS last updated on 18/Jun/20

$$\mathrm{the}\mathrm{colour}\mathrm{is}\mathrm{nice}\mathrm{but}\mathrm{hard}\mathrm{to}\mathrm{read}...$$

Commented by Rasheed.Sindhi last updated on 18/Jun/20

$$\mathcal{T}hen{it}^{\prime }snotnicehere!Main$$$$purposeistoberead.{Isn}^{\prime }tit?:)$$

Answered by mathmax by abdo last updated on 18/Jun/20

$$\mathrm{b})\mathrm{I}=\int \frac{{\mathrm{x}}^2+\mathrm{x}+1}{\sqrt{1-{\mathrm{x}}^2}}\mathrm{dx}\mathrm{changement}\mathrm{x}=\mathrm{sint}\mathrm{give}$$$$\mathrm{I}=\int \frac{{\sin }^2\mathrm{t}+\mathrm{sint}+1}{\mathrm{cost}}\mathrm{cost}\mathrm{dt}=\int ({\sin }^2\mathrm{t}+\mathrm{sint}+1)\mathrm{dt}$$$$=\int \frac{1-\cos \left(2\mathrm{t}\right)}{2}\mathrm{dt}-\mathrm{cost}+\mathrm{t}$$$$=\frac{\mathrm{t}}{2}-\frac{1}{4}\sin \left(2\mathrm{t}\right)-\mathrm{cost}+\mathrm{t}+\mathrm{c}$$$$=\frac{3}{2}\mathrm{t}-\frac{1}{2}\mathrm{sint}\mathrm{cost}-\mathrm{cost}+\mathrm{c}=\frac{3}{2}\mathrm{arcsinx}-\frac{\mathrm{x}}{2}\sqrt{1-{\mathrm{x}}^2}-\sqrt{1-{\mathrm{x}}^2}+\mathrm{c}$$

Commented by Mr.D.N. last updated on 18/Jun/20

thanks to your pleasure time.✍��

Commented by mathmax by abdo last updated on 18/Jun/20

$$\mathrm{you}\mathrm{are}\mathrm{welcome}.$$

Answered by mathmax by abdo last updated on 18/Jun/20

$$\mathrm{a})\mathrm{at}\mathrm{form}\mathrm{of}\mathrm{serie}\mathrm{I}=\int \frac{{\mathrm{xe}}^{\mathrm{x}}}{{(\mathrm{x}+1)}^2}\mathrm{dx}\mathrm{changement}\mathrm{x}+1=\mathrm{t}\mathrm{give}$$$$\mathrm{I}=\int \frac{(\mathrm{t}-1){\mathrm{e}}^{\mathrm{t}-1}}{{\mathrm{t}}^2}\mathrm{dt}=\frac{1}{\mathrm{e}}\int \frac{(\mathrm{t}-1)}{{\mathrm{t}}^2}{\mathrm{e}}^{\mathrm{t}}\mathrm{dt}$$$$=\frac{1}{\mathrm{e}}\int (\frac{1}{\mathrm{t}}-\frac{1}{{\mathrm{t}}^2})\sum _{\mathrm{n}=0}^{\mathrm{\infty }}\frac{{\mathrm{t}}^{\mathrm{n}}}{\mathrm{n}!}\mathrm{dt}=\frac{1}{\mathrm{e}}\sum _{\mathrm{n}=0}^{\mathrm{\infty }}\frac{1}{\mathrm{n}!}\int ({\mathrm{t}}^{\mathrm{n}-1}-{\mathrm{t}}^{\mathrm{n}-2})\mathrm{dt}$$$$=\frac{1}{\mathrm{e}}\sum _{\mathrm{n}=2}^{\mathrm{\infty }}\frac{1}{\mathrm{n}!}(\frac{{\mathrm{t}}^{\mathrm{n}}}{\mathrm{n}}-\frac{{\mathrm{t}}^{\mathrm{n}-1}}{\mathrm{n}-1})+\frac{1}{\mathrm{e}}\int (\frac{1}{\mathrm{t}}-\frac{1}{{\mathrm{t}}^2})\mathrm{dt}+\frac{1}{\mathrm{e}}\int (1-\frac{1}{\mathrm{t}})\mathrm{dt}$$$$=\frac{1}{\mathrm{e}}\sum _{\mathrm{n}=2}^{\mathrm{\infty }}\frac{{\mathrm{t}}^{\mathrm{n}}}{\mathrm{n}(\mathrm{n}!)}-\frac{1}{\mathrm{e}}\sum _{\mathrm{n}=2}^{\mathrm{\infty }}\frac{{\mathrm{t}}^{\mathrm{n}-1}}{(\mathrm{n}-1)\mathrm{n}!}+\frac{\ln \mid \mathrm{t}\mid }{\mathrm{e}}+\frac{1}{\mathrm{et}}+\frac{\mathrm{t}}{\mathrm{e}}-\frac{\ln \mid \mathrm{t}\mid }{\mathrm{e}}$$$$=\frac{1}{\mathrm{e}}\sum _{\mathrm{n}=2}^{\mathrm{\infty }}\frac{{(\mathrm{x}+1)}^{\mathrm{n}}}{\mathrm{n}(\mathrm{n}!)}-\frac{1}{\mathrm{e}}\sum _{\mathrm{n}=2}^{\mathrm{\infty }}\frac{{(\mathrm{x}+1)}^{\mathrm{n}-1}}{(\mathrm{n}-1)\mathrm{n}!}+\frac{1}{\mathrm{e}(\mathrm{x}+1)}+\frac{\mathrm{x}+1}{\mathrm{e}}$$

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