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Question Number 84809 by M±th+et£s last updated on 16/Mar/20

∫(x/((x^2 +1)^(3/2) arctan(x))) dx

$$\int\frac{{x}}{\left({x}^{\mathrm{2}} +\mathrm{1}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} {arctan}\left({x}\right)}\:{dx} \\ $$

Commented by abdomathmax last updated on 18/Mar/20

A =∫ (x/((x^(2 ) +1)^(3/2) arctanx)) changement arctanx=t give A =∫ ((tant)/((1+tan^2 t)^(3/2) t))×(1+tan^2 t)dt =∫ ((tant)/(t(√(1+tan^2 t))))dt =∫ ((sint)/(tcost))×cost dt =∫ ((sint)/t)dt

$${A}\:=\int\:\:\:\:\frac{{x}}{\left({x}^{\mathrm{2}\:} +\mathrm{1}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} \:{arctanx}}\:{changement}\:{arctanx}={t} \\ $$$${give}\:{A}\:=\int\:\:\frac{{tant}}{\left(\mathrm{1}+{tan}^{\mathrm{2}} {t}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} {t}}×\left(\mathrm{1}+{tan}^{\mathrm{2}} {t}\right){dt} \\ $$$$=\int\:\:\frac{{tant}}{{t}\sqrt{\mathrm{1}+{tan}^{\mathrm{2}} {t}}}{dt}\:=\int\:\frac{{sint}}{{tcost}}×{cost}\:{dt}\:=\int\:\frac{{sint}}{{t}}{dt} \\ $$

Answered by TANMAY PANACEA last updated on 16/Mar/20

x=tana ∫((tana.sec^2 a)/(sec^3 a.a))da ∫((sina)/(cosa.seca.a))da ∫((sina)/a)da ∫((a−(a^3 /(3!))+(a^5 /(5!))−(a^7 /(7!))+...)/a) da ∫1−(a^2 /(3!))+(a^4 /(5!))−(a^6 /(7!))+... a−(a^3 /(3×3!))+(a^5 /(5×5!))−(a^7 /(7×7!))+...∞ =Σ_(n=1) ^∞ (a^(2n−1) /((2n−1).(2n−1)!)) =Σ_(n=1) ^∞ (((tan^(−1) x)^(2n−1) )/((2n−1)(2n−1)!))

$${x}={tana} \\ $$$$\int\frac{{tana}.{sec}^{\mathrm{2}} {a}}{{sec}^{\mathrm{3}} {a}.{a}}{da} \\ $$$$\int\frac{{sina}}{{cosa}.{seca}.{a}}{da} \\ $$$$\int\frac{{sina}}{{a}}{da} \\ $$$$\int\frac{{a}−\frac{{a}^{\mathrm{3}} }{\mathrm{3}!}+\frac{{a}^{\mathrm{5}} }{\mathrm{5}!}−\frac{{a}^{\mathrm{7}} }{\mathrm{7}!}+...}{{a}}\:{da} \\ $$$$\int\mathrm{1}−\frac{{a}^{\mathrm{2}} }{\mathrm{3}!}+\frac{{a}^{\mathrm{4}} }{\mathrm{5}!}−\frac{{a}^{\mathrm{6}} }{\mathrm{7}!}+... \\ $$$${a}−\frac{{a}^{\mathrm{3}} }{\mathrm{3}×\mathrm{3}!}+\frac{{a}^{\mathrm{5}} }{\mathrm{5}×\mathrm{5}!}−\frac{{a}^{\mathrm{7}} }{\mathrm{7}×\mathrm{7}!}+...\infty \\ $$$$=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{a}^{\mathrm{2}{n}−\mathrm{1}} }{\left(\mathrm{2}{n}−\mathrm{1}\right).\left(\mathrm{2}{n}−\mathrm{1}\right)!} \\ $$$$=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left({tan}^{−\mathrm{1}} {x}\right)^{\mathrm{2}{n}−\mathrm{1}} }{\left(\mathrm{2}{n}−\mathrm{1}\right)\left(\mathrm{2}{n}−\mathrm{1}\right)!} \\ $$

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