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Question Number 156779 by Fresnel last updated on 15/Oct/21

∫((ln(1+x^2 ))/(1+x^2 ))

$$\int\frac{{ln}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)}{\mathrm{1}+{x}^{\mathrm{2}} } \\ $$

Answered by phanphuoc last updated on 15/Oct/21

x=tant

$${x}={tant} \\ $$

Answered by puissant last updated on 15/Oct/21

Q=∫((ln(1+x^2 ))/(1+x^2 ))dx=∫Σ_(n≥1) (((−1)^(n+1) )/n)•(x^(2n) /(1+x^2 ))dx =Σ_(n≥1) (((−1)^(n+1) )/n)∫((x^(2n) −x^(4n) )/(1−x^4 ))dx ; x=u^(1/4) →dx=(1/4)u^(−(3/4)) du ⇒ Q=Σ_(n≥1) (((−1)^(n+1) )/(4n))∫((u^((8n−3)/4) −u^((16n−3)/4) )/(1−u))du ⇒ Q=Σ_(n≥1) (((−1)^(n+1) )/(4n)){ψ(((16n−3)/4)+1)−ψ(((8n−3)/4)+1)} ∵∴ Q= Σ_(n≥1) (((−1)^(n+1) )/(4n)){ψ(((16n+1)/4))−ψ(((8n+1)/4))} .............Le puissant............

$${Q}=\int\frac{{ln}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)}{\mathrm{1}+{x}^{\mathrm{2}} }{dx}=\int\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{\left(−\mathrm{1}\right)^{{n}+\mathrm{1}} }{{n}}\bullet\frac{{x}^{\mathrm{2}{n}} }{\mathrm{1}+{x}^{\mathrm{2}} }{dx} \\ $$$$=\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{\left(−\mathrm{1}\right)^{{n}+\mathrm{1}} }{{n}}\int\frac{{x}^{\mathrm{2}{n}} −{x}^{\mathrm{4}{n}} }{\mathrm{1}−{x}^{\mathrm{4}} }{dx}\:;\:{x}={u}^{\frac{\mathrm{1}}{\mathrm{4}}} \rightarrow{dx}=\frac{\mathrm{1}}{\mathrm{4}}{u}^{−\frac{\mathrm{3}}{\mathrm{4}}} {du} \\ $$$$\Rightarrow\:{Q}=\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{\left(−\mathrm{1}\right)^{{n}+\mathrm{1}} }{\mathrm{4}{n}}\int\frac{{u}^{\frac{\mathrm{8}{n}−\mathrm{3}}{\mathrm{4}}} −{u}^{\frac{\mathrm{16}{n}−\mathrm{3}}{\mathrm{4}}} \:}{\mathrm{1}−{u}}{du} \\ $$$$\Rightarrow\:{Q}=\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{\left(−\mathrm{1}\right)^{{n}+\mathrm{1}} }{\mathrm{4}{n}}\left\{\psi\left(\frac{\mathrm{16}{n}−\mathrm{3}}{\mathrm{4}}+\mathrm{1}\right)−\psi\left(\frac{\mathrm{8}{n}−\mathrm{3}}{\mathrm{4}}+\mathrm{1}\right)\right\} \\ $$$$\:\:\:\:\:\:\:\:\because\therefore\:\:{Q}=\:\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{\left(−\mathrm{1}\right)^{{n}+\mathrm{1}} }{\mathrm{4}{n}}\left\{\psi\left(\frac{\mathrm{16}{n}+\mathrm{1}}{\mathrm{4}}\right)−\psi\left(\frac{\mathrm{8}{n}+\mathrm{1}}{\mathrm{4}}\right)\right\} \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:.............\mathscr{L}{e}\:{puissant}............ \\ $$

Commented by Fresnel last updated on 15/Oct/21

Merci

$${Merci} \\ $$

Commented by Ruuudiy last updated on 17/Oct/21

Thanks S

$$\mathfrak{Thanks}\:\mathcal{S} \: \: \: \: \\ $$

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