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1a-dx-e-5x-e-3x-e-2x-1b-dx-e-5x-e-3x-e-2x-2a-dx-x-5-3-x-3-5-2b-dx-x-5-3-x-3-5-

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Question Number 46323 by MJS last updated on 24/Oct/18
(1a) ∫(dx/(e^(5x) +e^(3x) +e^(2x) ))=? (1b) ∫(dx/(e^(5x) −e^(3x) −e^(2x) ))=? (2a) ∫(dx/(x^(5/3) +x^(3/5) ))=? (2b) ∫(dx/(x^(5/3) −x^(3/5) ))=?
$$\left(\mathrm{1}{a}\right)\:\:\int\frac{{dx}}{\mathrm{e}^{\mathrm{5}{x}} +\mathrm{e}^{\mathrm{3}{x}} +\mathrm{e}^{\mathrm{2}{x}} }=? \\ $$$$\left(\mathrm{1}{b}\right)\:\:\int\frac{{dx}}{\mathrm{e}^{\mathrm{5}{x}} −\mathrm{e}^{\mathrm{3}{x}} −\mathrm{e}^{\mathrm{2}{x}} }=? \\ $$$$\left(\mathrm{2}{a}\right)\:\:\int\frac{{dx}}{{x}^{\frac{\mathrm{5}}{\mathrm{3}}} +{x}^{\frac{\mathrm{3}}{\mathrm{5}}} }=? \\ $$$$\left(\mathrm{2}{b}\right)\:\:\int\frac{{dx}}{{x}^{\frac{\mathrm{5}}{\mathrm{3}}} −{x}^{\frac{\mathrm{3}}{\mathrm{5}}} }=? \\ $$
Commented by maxmathsup by imad last updated on 24/Oct/18
(2a) we use the changement x=t^(15) ⇒ I = ∫ ((15 t^(14) dt)/((t^(15) )^(5/3) +(t^(15) )^(3/5) )) = ∫ ((15 t^(14) )/(t^(25) + t^9 )) dt =15 ∫ (t^5 /(t^(16) +1)) dt but t^(16) +1 =(t^8 )^2 +1 =(t^8 +1)^2 −2t^8 =(t^8 +1−(√2)t^4 )(t^8 +1+(√2)t^4 ) ⇒ (1/((t^8 −(√2)t^4 +1)(t^8 +(√2)t^4 +1))) =(1/λ){ (1/(t^8 +(√2)t^4 +1)) −(1/(t^8 −(√2)t^4 +1))} λ=t^8 −(√2)t^4 +1−t^8 −(√2)t^4 −1 = −2(√2)t^4 ⇒ F(x)=(t^5 /(t^(16) +1)) =−(1/(2(√2))){ (t/(t^8 +(√2)t^4 +1)) −(t/(t^8 −(√2)t^4 +1))} ⇒ I =−((15)/(2(√2))) { ∫ ((tdt)/(t^8 +(√2)t^4 +1)) −∫ ((tdt)/(t^8 −(√2) t^4 +1))} also t^8 +(√2)t^4 +1 =u^2 +(√2)u +1 (u=t^4 ) =u^2 +2 ((√2)/2)u +(1/2) +(1/2) =(u +((√2)/2))^2 +(1/2) =(t^4 +((√2)/2))^2 +(1/2) ⇒ ∫ ((t dt)/(t^8 +(√2)t^4 +1)) =∫ (t/((t^4 +((√2)/2))^2 +(1/2))) dt changement t^4 =α give t =α^(1/4) ∫ (....)dt = ∫ (α^(1/4) /((α+((√2)/2))^2 +(1/2))) (1/4) α^((1/4)−1) dα = ∫ (dα/( (√α){ (α+((√2)/2))^2 +(1/2)})) ....be continued....
$$\left(\mathrm{2}{a}\right)\:{we}\:{use}\:{the}\:{changement}\:{x}={t}^{\mathrm{15}} \:\Rightarrow \\ $$$${I}\:=\:\int\:\:\:\frac{\mathrm{15}\:{t}^{\mathrm{14}} {dt}}{\left({t}^{\mathrm{15}} \right)^{\frac{\mathrm{5}}{\mathrm{3}}} \:+\left({t}^{\mathrm{15}} \right)^{\frac{\mathrm{3}}{\mathrm{5}}} }\:=\:\int\:\:\:\frac{\mathrm{15}\:{t}^{\mathrm{14}} }{{t}^{\mathrm{25}} \:+\:{t}^{\mathrm{9}} }\:{dt}\:=\mathrm{15}\:\int\:\:\:\frac{{t}^{\mathrm{5}} }{{t}^{\mathrm{16}} \:+\mathrm{1}}\:{dt}\:\:{but} \\ $$$${t}^{\mathrm{16}} \:+\mathrm{1}\:=\left({t}^{\mathrm{8}} \right)^{\mathrm{2}} \:+\mathrm{1}\:=\left({t}^{\mathrm{8}} \:+\mathrm{1}\right)^{\mathrm{2}} \:−\mathrm{2}{t}^{\mathrm{8}} \:=\left({t}^{\mathrm{8}} \:+\mathrm{1}−\sqrt{\mathrm{2}}{t}^{\mathrm{4}} \right)\left({t}^{\mathrm{8}} \:+\mathrm{1}+\sqrt{\mathrm{2}}{t}^{\mathrm{4}} \right)\:\Rightarrow \\ $$$$\frac{\mathrm{1}}{\left({t}^{\mathrm{8}} \:−\sqrt{\mathrm{2}}{t}^{\mathrm{4}} \:+\mathrm{1}\right)\left({t}^{\mathrm{8}} \:+\sqrt{\mathrm{2}}{t}^{\mathrm{4}} \:+\mathrm{1}\right)}\:=\frac{\mathrm{1}}{\lambda}\left\{\:\frac{\mathrm{1}}{{t}^{\mathrm{8}} \:+\sqrt{\mathrm{2}}{t}^{\mathrm{4}} \:+\mathrm{1}}\:−\frac{\mathrm{1}}{{t}^{\mathrm{8}} \:−\sqrt{\mathrm{2}}{t}^{\mathrm{4}} \:+\mathrm{1}}\right\} \\ $$$$\lambda={t}^{\mathrm{8}} −\sqrt{\mathrm{2}}{t}^{\mathrm{4}} \:\:+\mathrm{1}−{t}^{\mathrm{8}} −\sqrt{\mathrm{2}}{t}^{\mathrm{4}} −\mathrm{1}\:=\:−\mathrm{2}\sqrt{\mathrm{2}}{t}^{\mathrm{4}} \:\Rightarrow \\ $$$${F}\left({x}\right)=\frac{{t}^{\mathrm{5}} }{{t}^{\mathrm{16}} \:+\mathrm{1}}\:=−\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}\left\{\:\frac{{t}}{{t}^{\mathrm{8}} \:+\sqrt{\mathrm{2}}{t}^{\mathrm{4}} \:+\mathrm{1}}\:−\frac{{t}}{{t}^{\mathrm{8}} \:−\sqrt{\mathrm{2}}{t}^{\mathrm{4}} \:+\mathrm{1}}\right\}\:\Rightarrow \\ $$$${I}\:=−\frac{\mathrm{15}}{\mathrm{2}\sqrt{\mathrm{2}}}\:\left\{\:\int\:\:\:\:\frac{{tdt}}{{t}^{\mathrm{8}} \:+\sqrt{\mathrm{2}}{t}^{\mathrm{4}} \:+\mathrm{1}}\:−\int\:\:\:\frac{{tdt}}{{t}^{\mathrm{8}} \:−\sqrt{\mathrm{2}}\:{t}^{\mathrm{4}} \:+\mathrm{1}}\right\}\:{also}\: \\ $$$${t}^{\mathrm{8}} \:+\sqrt{\mathrm{2}}{t}^{\mathrm{4}} \:+\mathrm{1}\:={u}^{\mathrm{2}} \:+\sqrt{\mathrm{2}}{u}\:+\mathrm{1}\:\:\left({u}={t}^{\mathrm{4}} \right) \\ $$$$={u}^{\mathrm{2}} \:+\mathrm{2}\:\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}{u}\:+\frac{\mathrm{1}}{\mathrm{2}}\:+\frac{\mathrm{1}}{\mathrm{2}}\:=\left({u}\:+\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\right)^{\mathrm{2}} \:+\frac{\mathrm{1}}{\mathrm{2}}\:=\left({t}^{\mathrm{4}} \:+\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\right)^{\mathrm{2}} \:+\frac{\mathrm{1}}{\mathrm{2}}\:\Rightarrow \\ $$$$\int\:\:\:\:\frac{{t}\:{dt}}{{t}^{\mathrm{8}} \:+\sqrt{\mathrm{2}}{t}^{\mathrm{4}} \:+\mathrm{1}}\:=\int\:\:\:\:\frac{{t}}{\left({t}^{\mathrm{4}} \:+\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\right)^{\mathrm{2}} \:+\frac{\mathrm{1}}{\mathrm{2}}}\:{dt}\:\:\:{changement}\:{t}^{\mathrm{4}} =\alpha\:{give}\:\:{t}\:=\alpha^{\frac{\mathrm{1}}{\mathrm{4}}} \\ $$$$\int\:\left(….\right){dt}\:=\:\int\:\:\:\:\frac{\alpha^{\frac{\mathrm{1}}{\mathrm{4}}} }{\left(\alpha+\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\right)^{\mathrm{2}} \:+\frac{\mathrm{1}}{\mathrm{2}}}\:\frac{\mathrm{1}}{\mathrm{4}}\:\alpha^{\frac{\mathrm{1}}{\mathrm{4}}−\mathrm{1}} \:{d}\alpha \\ $$$$=\:\int\:\:\:\:\:\:\:\:\frac{{d}\alpha}{\:\sqrt{\alpha}\left\{\:\:\left(\alpha+\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\right)^{\mathrm{2}} \:+\frac{\mathrm{1}}{\mathrm{2}}\right\}}\:….{be}\:{continued}…. \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 24/Oct/18
2a)∫(dx/(x^a +x^b )) [here ab=1 b=(5/3) a=(3/5)] ∫(dx/(x^b (x^(a−b) +1))) ∫x^(−b) (1−x^t +x^(2t) −x^(3t) +...)dx [ t=a−b = ∫{x^(−b) −x^(t−b) +x^(2t−b) −x^(3t−b) +.. }dx =(x^(−b+1) /(−b+1))−(x^(t−b+1) /(t−b+1))+(x^(2t−b+1) /(2t−b+1))−.... pls check may i proceed this way...
$$\left.\mathrm{2}{a}\right)\int\frac{{dx}}{{x}^{{a}} +{x}^{{b}} }\:\:\:\left[{here}\:{ab}=\mathrm{1}\:\:\:{b}=\frac{\mathrm{5}}{\mathrm{3}}\:\:\:{a}=\frac{\mathrm{3}}{\mathrm{5}}\right] \\ $$$$\int\frac{{dx}}{{x}^{{b}} \left({x}^{{a}−{b}} +\mathrm{1}\right)} \\ $$$$\int{x}^{−{b}} \left(\mathrm{1}−{x}^{{t}} +{x}^{\mathrm{2}{t}} −{x}^{\mathrm{3}{t}} +…\right){dx}\:\:\:\left[\:{t}={a}−{b}\:\:=\right. \\ $$$$\int\left\{{x}^{−{b}} −{x}^{{t}−{b}} +{x}^{\mathrm{2}{t}−{b}} −{x}^{\mathrm{3}{t}−{b}} +..\:\right\}{dx} \\ $$$$=\frac{{x}^{−{b}+\mathrm{1}} }{−{b}+\mathrm{1}}−\frac{{x}^{{t}−{b}+\mathrm{1}} }{{t}−{b}+\mathrm{1}}+\frac{{x}^{\mathrm{2}{t}−{b}+\mathrm{1}} }{\mathrm{2}{t}−{b}+\mathrm{1}}−…. \\ $$$${pls}\:{check}\:\:{may}\:{i}\:{proceed}\:{this}\:{way}… \\ $$$$ \\ $$$$ \\ $$$$\:\:\:\:\:\: \\ $$$$ \\ $$

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