Question Number 35049 by math khazana by abdo last updated on 14/May/18
$${let}\:{A}_{{n}} \:=\:\int_{\frac{\mathrm{1}}{{n}}} ^{{n}} \:\left(\mathrm{1}+\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right){arctanx}\:{dx} \\ $$$$\left.\mathrm{1}\right)\:{calculate}\:{A}_{{n}} \\ $$$$\left.\mathrm{2}\right)\:{find}\:{lim}_{{n}\rightarrow+\infty} \:{A}_{{n}} \\ $$
Commented by abdo mathsup 649 cc last updated on 16/May/18
![1) let integrate by parts u^′ =1+(1/x^2 ) and v=arctanx A_n = [(1−(1/x))arctanx]_(1/n) ^n −∫_(1/n) ^n (1−(1/x)) (dx/(1+x^2 )) =(1−(1/n))arctan(n) −(1−n)arctan((1/n)) − ∫_(1/n) ^n (dx/(1+x^2 )) + ∫_(1/n) ^n (dx/(x(1+x^2 ))) but ∫_(1/n) ^n (dx/(1+x^2 )) = arctan(n) −arctan((1/n)) ∫_(1/n) ^n (1/(x(1+x^2 ))) dx= ∫_(1/n) ^n ((1/x) −(x/(1+x^2 )))dx =ln(n) −ln((1/n)) −(1/2)[ln(1+x^2 )]_(1/n) ^n = 2ln(n) −(1/2){ ln(1+n^2 ) −ln(1+(1/n^2 ))}⇒ A_n = (1−(1/n))arctan(n)−(1−n)arctan((1/n)) −arctan(n) +arctan((1/n)) +2ln(n)−(1/2)(2ln(n)) A_n =−((arctan(n))/n) +n arctan((1/n)) +(π/2) −2arctan(n) +ln(n) 2) we have lim_(n→+∞) ( −((arctan(n))/n) +n arctan((1/n)) +(π/2) −2arctsn(n)) =0 +1 −(π/2) −π but lim_(n→+∞) ln(n) =+∞ ⇒ lim_(n→+∞) A_n = +∞.](https://www.tinkutara.com/question/Q35145.png)
$$\left.\mathrm{1}\right)\:{let}\:{integrate}\:{by}\:{parts}\:{u}^{'} \:=\mathrm{1}+\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\:\:{and}\:\:{v}={arctanx} \\ $$$${A}_{{n}} \:=\:\left[\left(\mathrm{1}−\frac{\mathrm{1}}{{x}}\right){arctanx}\right]_{\frac{\mathrm{1}}{{n}}} ^{{n}} \:\:−\int_{\frac{\mathrm{1}}{{n}}} ^{{n}} \:\:\left(\mathrm{1}−\frac{\mathrm{1}}{{x}}\right)\:\frac{{dx}}{\mathrm{1}+{x}^{\mathrm{2}} } \\ $$$$=\left(\mathrm{1}−\frac{\mathrm{1}}{{n}}\right){arctan}\left({n}\right)\:−\left(\mathrm{1}−{n}\right){arctan}\left(\frac{\mathrm{1}}{{n}}\right) \\ $$$$−\:\int_{\frac{\mathrm{1}}{{n}}} ^{{n}} \:\frac{{dx}}{\mathrm{1}+{x}^{\mathrm{2}} }\:\:+\:\int_{\frac{\mathrm{1}}{{n}}} ^{{n}} \:\:\:\frac{{dx}}{{x}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)}\:\:{but} \\ $$$$\int_{\frac{\mathrm{1}}{{n}}} ^{{n}} \:\:\:\frac{{dx}}{\mathrm{1}+{x}^{\mathrm{2}} }\:=\:{arctan}\left({n}\right)\:−{arctan}\left(\frac{\mathrm{1}}{{n}}\right) \\ $$$$\int_{\frac{\mathrm{1}}{{n}}} ^{{n}} \frac{\mathrm{1}}{{x}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)}\:{dx}=\:\int_{\frac{\mathrm{1}}{{n}}} ^{{n}} \left(\frac{\mathrm{1}}{{x}}\:−\frac{{x}}{\mathrm{1}+{x}^{\mathrm{2}} }\right){dx} \\ $$$$={ln}\left({n}\right)\:−{ln}\left(\frac{\mathrm{1}}{{n}}\right)\:\:−\frac{\mathrm{1}}{\mathrm{2}}\left[{ln}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)\right]_{\frac{\mathrm{1}}{{n}}} ^{{n}} \\ $$$$=\:\mathrm{2}{ln}\left({n}\right)\:−\frac{\mathrm{1}}{\mathrm{2}}\left\{\:{ln}\left(\mathrm{1}+{n}^{\mathrm{2}} \right)\:−{ln}\left(\mathrm{1}+\frac{\mathrm{1}}{{n}^{\mathrm{2}} }\right)\right\}\Rightarrow \\ $$$${A}_{{n}} \:=\:\left(\mathrm{1}−\frac{\mathrm{1}}{{n}}\right){arctan}\left({n}\right)−\left(\mathrm{1}−{n}\right){arctan}\left(\frac{\mathrm{1}}{{n}}\right) \\ $$$$−{arctan}\left({n}\right)\:+{arctan}\left(\frac{\mathrm{1}}{{n}}\right)\:\:+\mathrm{2}{ln}\left({n}\right)−\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{2}{ln}\left({n}\right)\right) \\ $$$${A}_{{n}} =−\frac{{arctan}\left({n}\right)}{{n}}\:+{n}\:{arctan}\left(\frac{\mathrm{1}}{{n}}\right)\:+\frac{\pi}{\mathrm{2}}\:−\mathrm{2}{arctan}\left({n}\right) \\ $$$$+{ln}\left({n}\right) \\ $$$$\left.\mathrm{2}\right)\:{we}\:{have} \\ $$$${lim}_{{n}\rightarrow+\infty} \left(\:−\frac{{arctan}\left({n}\right)}{{n}}\:+{n}\:{arctan}\left(\frac{\mathrm{1}}{{n}}\right)\:+\frac{\pi}{\mathrm{2}}\:−\mathrm{2}{arctsn}\left({n}\right)\right) \\ $$$$=\mathrm{0}\:+\mathrm{1}\:−\frac{\pi}{\mathrm{2}}\:−\pi\:\:\:{but}\:{lim}_{{n}\rightarrow+\infty} \:{ln}\left({n}\right)\:=+\infty\:\Rightarrow \\ $$$${lim}_{{n}\rightarrow+\infty} \:\:{A}_{{n}} =\:+\infty. \\ $$
Answered by MJS last updated on 15/May/18
![∫(1+(1/x^2 ))arctan(x)dx= =∫arctan(x)dx+∫((arctan(x))/x^2 )dx= ∫arctan(x)dx= [((∫u′v=uv−∫uv′)),((u′=1 ⇒ u=x)),((v=arctan(x) ⇒ v′=(1/(x^2 +1)))) ] =xarctan(x)−∫(x/(x^2 +1))dx= [((t=x^2 +1 → dx=(dt/(2x)))) ] =xarctan(x)−(1/2)∫(1/t)dt= =xarctan(x)−((ln(t))/2)= =xarctan(x)−((ln(x^2 +1))/2) ∫((arctan(x))/x^2 )dx= [((∫u′v=uv−∫uv′)),((u′=(1/x^2 ) ⇒ u=−(1/x))),((v=arctan(x) ⇒ v′=(1/(x^2 +1)))) ] =−((arctan(x))/x)+∫(1/(x(x^2 +1)))dx= =−((arctan(x))/x)+∫((1/x)−(x^2 /(x^2 +1)))dx= [((same as above)) ] =−((arctan(x))/x)+ln(x)−((ln(x^2 +1))/2) =xarctan(x)−((ln(x^2 +1))/2)−((arctan(x))/x)+ln(x)−((ln(x^2 +1))/2)= =(((x^2 −1)/x))arctan(x)+ln((x/(x^2 +1)))+C ∫_(1/n) ^n (1+(1/x^2 ))arctan(x)dx= =(((n^2 −1)/n))arctan(n)+ln((n/(n^2 +1)))−(((((1/n^2 )−1)/(1/n)))arctan((1/n))+ln(((1/n)/((1/n^2 )+1))))= =(((n^2 −1)/n))arctan(n)+ln((n/(n^2 +1)))−((((1−n^2 )/n))((π/2)−arctan(n))+ln((n/(n^2 +1))))= =((π(n^2 −1))/(2n)) lim_(n→∞) ((π(n^2 −1))/(2n))=(π/2)lim_(n→∞) (n−(1/n))=∞](https://www.tinkutara.com/question/Q35081.png)
$$\int\left(\mathrm{1}+\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right)\mathrm{arctan}\left({x}\right){dx}= \\ $$$$=\int\mathrm{arctan}\left({x}\right){dx}+\int\frac{\mathrm{arctan}\left({x}\right)}{{x}^{\mathrm{2}} }{dx}= \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\int\mathrm{arctan}\left({x}\right){dx}= \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\begin{bmatrix}{\int{u}'{v}={uv}−\int{uv}'}\\{{u}'=\mathrm{1}\:\Rightarrow\:{u}={x}}\\{{v}=\mathrm{arctan}\left({x}\right)\:\Rightarrow\:{v}'=\frac{\mathrm{1}}{{x}^{\mathrm{2}} +\mathrm{1}}}\end{bmatrix} \\ $$$$\:\:\:\:\:\:\:\:\:\:={x}\mathrm{arctan}\left({x}\right)−\int\frac{{x}}{{x}^{\mathrm{2}} +\mathrm{1}}{dx}= \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\begin{bmatrix}{{t}={x}^{\mathrm{2}} +\mathrm{1}\:\rightarrow\:{dx}=\frac{{dt}}{\mathrm{2}{x}}}\end{bmatrix} \\ $$$$\:\:\:\:\:\:\:\:\:\:={x}\mathrm{arctan}\left({x}\right)−\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\mathrm{1}}{{t}}{dt}= \\ $$$$\:\:\:\:\:\:\:\:\:\:={x}\mathrm{arctan}\left({x}\right)−\frac{\mathrm{ln}\left({t}\right)}{\mathrm{2}}= \\ $$$$\:\:\:\:\:\:\:\:\:\:={x}\mathrm{arctan}\left({x}\right)−\frac{\mathrm{ln}\left({x}^{\mathrm{2}} +\mathrm{1}\right)}{\mathrm{2}} \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\int\frac{\mathrm{arctan}\left({x}\right)}{{x}^{\mathrm{2}} }{dx}= \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\begin{bmatrix}{\int{u}'{v}={uv}−\int{uv}'}\\{{u}'=\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\:\Rightarrow\:{u}=−\frac{\mathrm{1}}{{x}}}\\{{v}=\mathrm{arctan}\left({x}\right)\:\Rightarrow\:{v}'=\frac{\mathrm{1}}{{x}^{\mathrm{2}} +\mathrm{1}}}\end{bmatrix} \\ $$$$\:\:\:\:\:\:\:\:\:\:=−\frac{\mathrm{arctan}\left({x}\right)}{{x}}+\int\frac{\mathrm{1}}{{x}\left({x}^{\mathrm{2}} +\mathrm{1}\right)}{dx}= \\ $$$$\:\:\:\:\:\:\:\:\:\:=−\frac{\mathrm{arctan}\left({x}\right)}{{x}}+\int\left(\frac{\mathrm{1}}{{x}}−\frac{{x}^{\mathrm{2}} }{{x}^{\mathrm{2}} +\mathrm{1}}\right){dx}= \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\begin{bmatrix}{\mathrm{same}\:\mathrm{as}\:\mathrm{above}}\end{bmatrix} \\ $$$$\:\:\:\:\:\:\:\:\:\:=−\frac{\mathrm{arctan}\left({x}\right)}{{x}}+\mathrm{ln}\left({x}\right)−\frac{\mathrm{ln}\left({x}^{\mathrm{2}} +\mathrm{1}\right)}{\mathrm{2}} \\ $$$$ \\ $$$$={x}\mathrm{arctan}\left({x}\right)−\frac{\mathrm{ln}\left({x}^{\mathrm{2}} +\mathrm{1}\right)}{\mathrm{2}}−\frac{\mathrm{arctan}\left({x}\right)}{{x}}+\mathrm{ln}\left({x}\right)−\frac{\mathrm{ln}\left({x}^{\mathrm{2}} +\mathrm{1}\right)}{\mathrm{2}}= \\ $$$$=\left(\frac{{x}^{\mathrm{2}} −\mathrm{1}}{{x}}\right)\mathrm{arctan}\left({x}\right)+\mathrm{ln}\left(\frac{{x}}{{x}^{\mathrm{2}} +\mathrm{1}}\right)+{C} \\ $$$$\underset{\frac{\mathrm{1}}{{n}}} {\overset{{n}} {\int}}\left(\mathrm{1}+\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right)\mathrm{arctan}\left({x}\right){dx}= \\ $$$$=\left(\frac{{n}^{\mathrm{2}} −\mathrm{1}}{{n}}\right)\mathrm{arctan}\left({n}\right)+\mathrm{ln}\left(\frac{{n}}{{n}^{\mathrm{2}} +\mathrm{1}}\right)−\left(\left(\frac{\frac{\mathrm{1}}{{n}^{\mathrm{2}} }−\mathrm{1}}{\frac{\mathrm{1}}{{n}}}\right)\mathrm{arctan}\left(\frac{\mathrm{1}}{{n}}\right)+\mathrm{ln}\left(\frac{\frac{\mathrm{1}}{{n}}}{\frac{\mathrm{1}}{{n}^{\mathrm{2}} }+\mathrm{1}}\right)\right)= \\ $$$$=\left(\frac{{n}^{\mathrm{2}} −\mathrm{1}}{{n}}\right)\mathrm{arctan}\left({n}\right)+\mathrm{ln}\left(\frac{{n}}{{n}^{\mathrm{2}} +\mathrm{1}}\right)−\left(\left(\frac{\mathrm{1}−{n}^{\mathrm{2}} }{{n}}\right)\left(\frac{\pi}{\mathrm{2}}−\mathrm{arctan}\left({n}\right)\right)+\mathrm{ln}\left(\frac{\mathrm{n}}{\mathrm{n}^{\mathrm{2}} +\mathrm{1}}\right)\right)= \\ $$$$=\frac{\pi\left({n}^{\mathrm{2}} −\mathrm{1}\right)}{\mathrm{2}{n}} \\ $$$$ \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\frac{\pi\left({n}^{\mathrm{2}} −\mathrm{1}\right)}{\mathrm{2}{n}}=\frac{\pi}{\mathrm{2}}\underset{{n}\rightarrow\infty} {\mathrm{lim}}\left({n}−\frac{\mathrm{1}}{{n}}\right)=\infty \\ $$