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Question-99077

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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
the colour is nice but hard to read...
$$\mathrm{the}\:\mathrm{colour}\:\mathrm{is}\:\mathrm{nice}\:\mathrm{but}\:\mathrm{hard}\:\mathrm{to}\:\mathrm{read}… \\ $$
Commented by Rasheed.Sindhi last updated on 18/Jun/20
Then it′s not nice here! Main purpose is to be read. Isn′t it? :)
$$\mathcal{T}{hen}\:{it}'{s}\:{not}\:{nice}\:{here}!\:{Main} \\ $$$$\left.{purpose}\:{is}\:{to}\:{be}\:{read}.\:{Isn}'{t}\:{it}?\:\::\right) \\ $$
Answered by mathmax by abdo last updated on 18/Jun/20
b) I =∫((x^2 +x+1)/( (√(1−x^2 )))) dx changement x =sint give I =∫ ((sin^2 t +sint +1)/(cost))cost dt =∫ (sin^2 t+sint +1)dt =∫ ((1−cos(2t))/2)dt −cost +t =(t/2) −(1/4)sin(2t)−cost +t +c =(3/2)t −(1/2)sint cost −cost +c =(3/2) arcsinx −(x/2)(√(1−x^2 )) −(√(1−x^2 )) +c
$$\left.\mathrm{b}\right)\:\mathrm{I}\:=\int\frac{\mathrm{x}^{\mathrm{2}} \:+\mathrm{x}+\mathrm{1}}{\:\sqrt{\mathrm{1}−\mathrm{x}^{\mathrm{2}} }}\:\mathrm{dx}\:\mathrm{changement}\:\mathrm{x}\:=\mathrm{sint}\:\mathrm{give} \\ $$$$\mathrm{I}\:=\int\:\frac{\mathrm{sin}^{\mathrm{2}} \mathrm{t}\:+\mathrm{sint}\:+\mathrm{1}}{\mathrm{cost}}\mathrm{cost}\:\mathrm{dt}\:=\int\:\left(\mathrm{sin}^{\mathrm{2}} \mathrm{t}+\mathrm{sint}\:+\mathrm{1}\right)\mathrm{dt} \\ $$$$=\int\:\frac{\mathrm{1}−\mathrm{cos}\left(\mathrm{2t}\right)}{\mathrm{2}}\mathrm{dt}\:\:−\mathrm{cost}\:+\mathrm{t} \\ $$$$=\frac{\mathrm{t}}{\mathrm{2}}\:−\frac{\mathrm{1}}{\mathrm{4}}\mathrm{sin}\left(\mathrm{2t}\right)−\mathrm{cost}\:+\mathrm{t}\:+\mathrm{c} \\ $$$$=\frac{\mathrm{3}}{\mathrm{2}}\mathrm{t}\:−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sint}\:\mathrm{cost}\:−\mathrm{cost}\:+\mathrm{c}\:=\frac{\mathrm{3}}{\mathrm{2}}\:\mathrm{arcsinx}\:−\frac{\mathrm{x}}{\mathrm{2}}\sqrt{\mathrm{1}−\mathrm{x}^{\mathrm{2}} }\:−\sqrt{\mathrm{1}−\mathrm{x}^{\mathrm{2}} }\:+\mathrm{c} \\ $$
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
you are welcome .
$$\mathrm{you}\:\mathrm{are}\:\mathrm{welcome}\:. \\ $$
Answered by mathmax by abdo last updated on 18/Jun/20
a) at form of serie I =∫ ((xe^x )/((x+1)^2 ))dx changement x+1 =t give I =∫ (((t−1)e^(t−1) )/t^2 )dt =(1/e) ∫(((t−1))/t^2 ) e^t dt =(1/e) ∫ ((1/t)−(1/t^2 ))Σ_(n=0) ^∞ (t^n /(n!))dt =(1/e) Σ_(n=0) ^∞ (1/(n!))∫(t^(n−1) −t^(n−2) )dt =(1/e) Σ_(n=2) ^∞ (1/(n!))((t^n /n)−(t^(n−1) /(n−1))) +(1/e)∫ ((1/t)−(1/t^2 ))dt +(1/e)∫(1−(1/t))dt =(1/e) Σ_(n=2) ^∞ (t^n /(n(n!)))−(1/e)Σ_(n=2) ^∞ (t^(n−1) /((n−1)n!)) +((ln∣t∣)/e) +(1/(et)) +(t/e)−((ln∣t∣)/e) =(1/e) Σ_(n=2) ^∞ (((x+1)^n )/(n(n!))) −(1/e) Σ_(n=2) ^∞ (((x+1)^(n−1) )/((n−1)n!)) +(1/(e(x+1))) +((x+1)/e)
$$\left.\mathrm{a}\right)\:\mathrm{at}\:\mathrm{form}\:\mathrm{of}\:\mathrm{serie}\:\:\mathrm{I}\:=\int\:\frac{\mathrm{xe}^{\mathrm{x}} }{\left(\mathrm{x}+\mathrm{1}\right)^{\mathrm{2}} }\mathrm{dx}\:\:\mathrm{changement}\:\mathrm{x}+\mathrm{1}\:=\mathrm{t}\:\mathrm{give} \\ $$$$\mathrm{I}\:=\int\:\frac{\left(\mathrm{t}−\mathrm{1}\right)\mathrm{e}^{\mathrm{t}−\mathrm{1}} }{\mathrm{t}^{\mathrm{2}} }\mathrm{dt}\:=\frac{\mathrm{1}}{\mathrm{e}}\:\int\frac{\left(\mathrm{t}−\mathrm{1}\right)}{\mathrm{t}^{\mathrm{2}} }\:\mathrm{e}^{\mathrm{t}} \:\mathrm{dt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{e}}\:\int\:\left(\frac{\mathrm{1}}{\mathrm{t}}−\frac{\mathrm{1}}{\mathrm{t}^{\mathrm{2}} }\right)\sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \:\frac{\mathrm{t}^{\mathrm{n}} }{\mathrm{n}!}\mathrm{dt}\:=\frac{\mathrm{1}}{\mathrm{e}}\:\sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \:\frac{\mathrm{1}}{\mathrm{n}!}\int\left(\mathrm{t}^{\mathrm{n}−\mathrm{1}} −\mathrm{t}^{\mathrm{n}−\mathrm{2}} \right)\mathrm{dt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{e}}\:\sum_{\mathrm{n}=\mathrm{2}} ^{\infty} \:\frac{\mathrm{1}}{\mathrm{n}!}\left(\frac{\mathrm{t}^{\mathrm{n}} }{\mathrm{n}}−\frac{\mathrm{t}^{\mathrm{n}−\mathrm{1}} }{\mathrm{n}−\mathrm{1}}\right)\:+\frac{\mathrm{1}}{\mathrm{e}}\int\:\left(\frac{\mathrm{1}}{\mathrm{t}}−\frac{\mathrm{1}}{\mathrm{t}^{\mathrm{2}} }\right)\mathrm{dt}\:+\frac{\mathrm{1}}{\mathrm{e}}\int\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{t}}\right)\mathrm{dt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{e}}\:\sum_{\mathrm{n}=\mathrm{2}} ^{\infty} \:\frac{\mathrm{t}^{\mathrm{n}} }{\mathrm{n}\left(\mathrm{n}!\right)}−\frac{\mathrm{1}}{\mathrm{e}}\sum_{\mathrm{n}=\mathrm{2}} ^{\infty} \:\frac{\mathrm{t}^{\mathrm{n}−\mathrm{1}} }{\left(\mathrm{n}−\mathrm{1}\right)\mathrm{n}!}\:+\frac{\mathrm{ln}\mid\mathrm{t}\mid}{\mathrm{e}}\:+\frac{\mathrm{1}}{\mathrm{et}}\:+\frac{\mathrm{t}}{\mathrm{e}}−\frac{\mathrm{ln}\mid\mathrm{t}\mid}{\mathrm{e}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{e}}\:\sum_{\mathrm{n}=\mathrm{2}} ^{\infty} \:\frac{\left(\mathrm{x}+\mathrm{1}\right)^{\mathrm{n}} }{\mathrm{n}\left(\mathrm{n}!\right)}\:−\frac{\mathrm{1}}{\mathrm{e}}\:\sum_{\mathrm{n}=\mathrm{2}} ^{\infty} \:\frac{\left(\mathrm{x}+\mathrm{1}\right)^{\mathrm{n}−\mathrm{1}} }{\left(\mathrm{n}−\mathrm{1}\right)\mathrm{n}!}\:+\frac{\mathrm{1}}{\mathrm{e}\left(\mathrm{x}+\mathrm{1}\right)}\:+\frac{\mathrm{x}+\mathrm{1}}{\mathrm{e}} \\ $$

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