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Question Number 151780 by Tawa11 last updated on 23/Aug/21

Answered by puissant last updated on 23/Aug/21

Q=∫_0 ^1 ((arctanx)/x)dx arctanx=Σ_(n=0) ^∞ (−1)^n (x^(2n+1) /(2n+1)) ⇒ ((arctanx)/x)=Σ_(n=0) ^∞ (((−1)^n x^(2n) )/(2n+1)) ⇒ ∫_0 ^1 ((arctanx)/x)dx=Σ_(n=0) ^∞ (((−1)^n )/(2n+1))∫_0 ^1 x^(2n) dx = Σ_(n=0) ^∞ (((−1)^n )/((2n+1)^2 ))

$${Q}=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{arctanx}}{{x}}{dx} \\ $$$${arctanx}=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{n}} \frac{{x}^{\mathrm{2}{n}+\mathrm{1}} }{\mathrm{2}{n}+\mathrm{1}} \\ $$$$\Rightarrow\:\frac{{arctanx}}{{x}}=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} {x}^{\mathrm{2}{n}} }{\mathrm{2}{n}+\mathrm{1}} \\ $$$$\Rightarrow\:\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{arctanx}}{{x}}{dx}=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} }{\mathrm{2}{n}+\mathrm{1}}\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{\mathrm{2}{n}} {dx} \\ $$$$=\:\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} }{\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$

Commented by Tawa11 last updated on 23/Aug/21

Thanks sir. God bless you.

$$\mathrm{Thanks}\:\mathrm{sir}.\:\mathrm{God}\:\mathrm{bless}\:\mathrm{you}. \\ $$

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