Menu Close

find-x-arctan-x-1-x-dx-

Tinku Tara 0 Comments
Pin




Question Number 37335 by math khazana by abdo last updated on 12/Jun/18
find ∫ x arctan(x+(1/x))dx .
$${find}\:\int\:\:\:\:\:{x}\:{arctan}\left({x}+\frac{\mathrm{1}}{{x}}\right){dx}\:. \\ $$
Commented by math khazana by abdo last updated on 14/Jun/18
let I =∫ x arctan(x+(1/x))dx by parts u^′ =x and v=arctan(x+(1/x)) ⇒ I = (x^2 /2) arctan(x +(1/x)) −∫ (x^2 /2) (((1−(1/x^2 )))/(1+(x+(1/x))^2 ))dx =(x^2 /2)arctan(x+(1/x)) −(1/2)∫ ((x^2 −1)/(x^2 +(x^2 +1)^2 ))x^2 dx but ∫ ((x^4 −x^2 )/(x^2 +(x^2 +1)^2 ))dx = ∫ ((x^4 −x^2 )/(x^2 +x^4 +2x^2 +1))dx =∫ (( x^4 −x^2 )/(x^4 +3x^2 +1))dx =∫ ((x^4 +3x^2 +1 −3x^2 −1−x^2 )/(x^4 +3x^2 +1)) dx =x − ∫ ((4x^2 +1)/(x^4 +3x^2 +1)) dx let decompose F(x)= ((4x^2 +1)/(x^(4 ) +3x^2 +1)) roots of p(x)=x^4 +3x^2 +1 =q(t^2 ) t^2 +3t +1 Δ =9−4=5 ⇒ t_1 = ((−3 +(√5))/2) and t_2 = ((−3−(√5))/2) x^4 +3x^2 +1 =(x^2 −((−3+(√5))/2))( x^2 +((3+(√5))/2)) =(x^2 +((3−(√5))/2))( x^2 +((3+(√5))/2)) ⇒ F(x) = ((ax+b)/(x^2 +((3−(√5))/2))) +((cx+d)/(x^2 +((3+(√5))/2))) and ∫ F(x)dx =∫ ((ax+b)/(x^2 +((3−(√5))/2)))dx+ ∫ ((cx+d)/(x^2 +((3+(√5))/2)))dx.....
$${let}\:{I}\:=\int\:{x}\:{arctan}\left({x}+\frac{\mathrm{1}}{{x}}\right){dx}\:{by}\:{parts} \\ $$$${u}^{'} ={x}\:{and}\:{v}={arctan}\left({x}+\frac{\mathrm{1}}{{x}}\right)\:\Rightarrow \\ $$$${I}\:=\:\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\:{arctan}\left({x}\:+\frac{\mathrm{1}}{{x}}\right)\:−\int\:\:\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\:\frac{\left(\mathrm{1}−\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right)}{\mathrm{1}+\left({x}+\frac{\mathrm{1}}{{x}}\right)^{\mathrm{2}} }{dx} \\ $$$$=\frac{{x}^{\mathrm{2}} }{\mathrm{2}}{arctan}\left({x}+\frac{\mathrm{1}}{{x}}\right)\:−\frac{\mathrm{1}}{\mathrm{2}}\int\:\:\:\:\frac{{x}^{\mathrm{2}} \:−\mathrm{1}}{{x}^{\mathrm{2}} \:+\left({x}^{\mathrm{2}} \:+\mathrm{1}\right)^{\mathrm{2}} }{x}^{\mathrm{2}} \:{dx} \\ $$$${but}\:\:\int\:\:\:\:\:\:\frac{{x}^{\mathrm{4}} \:−{x}^{\mathrm{2}} }{{x}^{\mathrm{2}} \:+\left({x}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} }{dx} \\ $$$$=\:\int\:\:\:\:\:\frac{{x}^{\mathrm{4}} \:−{x}^{\mathrm{2}} }{{x}^{\mathrm{2}} \:+{x}^{\mathrm{4}} \:+\mathrm{2}{x}^{\mathrm{2}} \:+\mathrm{1}}{dx} \\ $$$$=\int\:\:\:\:\frac{\:{x}^{\mathrm{4}} \:−{x}^{\mathrm{2}} }{{x}^{\mathrm{4}} \:+\mathrm{3}{x}^{\mathrm{2}} \:+\mathrm{1}}{dx} \\ $$$$=\int\:\:\:\frac{{x}^{\mathrm{4}} \:+\mathrm{3}{x}^{\mathrm{2}} \:+\mathrm{1}\:−\mathrm{3}{x}^{\mathrm{2}} −\mathrm{1}−{x}^{\mathrm{2}} }{{x}^{\mathrm{4}} \:+\mathrm{3}{x}^{\mathrm{2}} \:+\mathrm{1}}\:{dx} \\ $$$$={x}\:\:−\:\int\:\:\:\:\frac{\mathrm{4}{x}^{\mathrm{2}} \:+\mathrm{1}}{{x}^{\mathrm{4}} \:+\mathrm{3}{x}^{\mathrm{2}} \:+\mathrm{1}}\:{dx}\:\:{let}\:{decompose} \\ $$$${F}\left({x}\right)=\:\frac{\mathrm{4}{x}^{\mathrm{2}} \:+\mathrm{1}}{{x}^{\mathrm{4}\:} \:+\mathrm{3}{x}^{\mathrm{2}} \:+\mathrm{1}} \\ $$$${roots}\:{of}\:{p}\left({x}\right)={x}^{\mathrm{4}} \:+\mathrm{3}{x}^{\mathrm{2}} \:+\mathrm{1}\:={q}\left({t}^{\mathrm{2}} \right) \\ $$$${t}^{\mathrm{2}} \:+\mathrm{3}{t}\:+\mathrm{1}\:\:\Delta\:=\mathrm{9}−\mathrm{4}=\mathrm{5}\:\Rightarrow \\ $$$${t}_{\mathrm{1}} =\:\frac{−\mathrm{3}\:+\sqrt{\mathrm{5}}}{\mathrm{2}}\:\:{and}\:\:{t}_{\mathrm{2}} \:=\:\frac{−\mathrm{3}−\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$$${x}^{\mathrm{4}} \:+\mathrm{3}{x}^{\mathrm{2}} \:+\mathrm{1}\:=\left({x}^{\mathrm{2}} \:−\frac{−\mathrm{3}+\sqrt{\mathrm{5}}}{\mathrm{2}}\right)\left(\:{x}^{\mathrm{2}} \:+\frac{\mathrm{3}+\sqrt{\mathrm{5}}}{\mathrm{2}}\right) \\ $$$$=\left({x}^{\mathrm{2}} \:\:+\frac{\mathrm{3}−\sqrt{\mathrm{5}}}{\mathrm{2}}\right)\left(\:{x}^{\mathrm{2}} \:+\frac{\mathrm{3}+\sqrt{\mathrm{5}}}{\mathrm{2}}\right)\:\Rightarrow \\ $$$${F}\left({x}\right)\:=\:\frac{{ax}+{b}}{{x}^{\mathrm{2}} \:+\frac{\mathrm{3}−\sqrt{\mathrm{5}}}{\mathrm{2}}}\:\:+\frac{{cx}+{d}}{{x}^{\mathrm{2}} \:+\frac{\mathrm{3}+\sqrt{\mathrm{5}}}{\mathrm{2}}}\:\:{and} \\ $$$$\int\:{F}\left({x}\right){dx}\:=\int\:\:\:\:\frac{{ax}+{b}}{{x}^{\mathrm{2}} \:+\frac{\mathrm{3}−\sqrt{\mathrm{5}}}{\mathrm{2}}}{dx}+\:\int\:\frac{{cx}+{d}}{{x}^{\mathrm{2}} \:+\frac{\mathrm{3}+\sqrt{\mathrm{5}}}{\mathrm{2}}}{dx}….. \\ $$

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *