Menu Close

t-2-1-t-2-2-dx-

Tinku Tara 0 Comments
Pin




Question Number 93484 by mashallah last updated on 13/May/20
∫t^2 /(1+t^2 )^2 dx=
$$\int\mathrm{t}^{\mathrm{2}} /\left(\mathrm{1}+\mathrm{t}^{\mathrm{2}} \right)^{\mathrm{2}} \mathrm{dx}= \\ $$
Commented by prakash jain last updated on 13/May/20
function is in t so constant wrt x (t^2 /((1+t^2 )^2 ))x+C
$$\mathrm{function}\:\mathrm{is}\:\mathrm{in}\:{t}\:\mathrm{so}\:\mathrm{constant}\:\mathrm{wrt}\:{x} \\ $$$$\frac{{t}^{\mathrm{2}} }{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)^{\mathrm{2}} }{x}+{C} \\ $$
Commented by mathmax by abdo last updated on 13/May/20
I =∫ (t^2 /((1+t^2 )^2 ))dt ⇒I =∫ ((t^2 +1−1)/((1+t^2 )))dt =∫ (dt/(1+t^2 ))−∫ (dt/((1+t^2 )^2 )) we have ∫ (dt/(1+t^2 )) =arctan(t)+c_1 ∫ (dt/((1+t^2 )^2 )) =_(t=tanx) ∫ (((1+tan^2 x)dx)/((1+tan^2 x)^2 )) =∫cos^2 x dx =(1/2)∫(1+cos(2x))dx =(x/2) +(1/4)sin(2x) +c_2 sin(2x) =((2tanx)/(1+tan^2 x)) =((2t)/(1+t^2 )) ⇒∫(dt/((1+t^2 )^2 )) =(1/2)arctan(t) +(t/(2(1+t^2 ))) +c_2 ⇒I =(1/2)arctan(t)−(t/(2(1+t^2 ))) +C
$${I}\:=\int\:\:\frac{{t}^{\mathrm{2}} }{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)^{\mathrm{2}} }{dt}\:\Rightarrow{I}\:=\int\:\frac{{t}^{\mathrm{2}} +\mathrm{1}−\mathrm{1}}{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)}{dt}\:=\int\:\:\frac{{dt}}{\mathrm{1}+{t}^{\mathrm{2}} }−\int\:\:\frac{{dt}}{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)^{\mathrm{2}} } \\ $$$${we}\:{have}\:\int\:\:\frac{{dt}}{\mathrm{1}+{t}^{\mathrm{2}} }\:={arctan}\left({t}\right)+{c}_{\mathrm{1}} \\ $$$$\int\:\:\:\frac{{dt}}{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)^{\mathrm{2}} }\:=_{{t}={tanx}} \:\:\:\int\:\:\:\frac{\left(\mathrm{1}+{tan}^{\mathrm{2}} {x}\right){dx}}{\left(\mathrm{1}+{tan}^{\mathrm{2}} {x}\right)^{\mathrm{2}} }\:\:\:\:=\int{cos}^{\mathrm{2}} {x}\:{dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int\left(\mathrm{1}+{cos}\left(\mathrm{2}{x}\right)\right){dx}\:=\frac{{x}}{\mathrm{2}}\:+\frac{\mathrm{1}}{\mathrm{4}}{sin}\left(\mathrm{2}{x}\right)\:+{c}_{\mathrm{2}} \\ $$$${sin}\left(\mathrm{2}{x}\right)\:=\frac{\mathrm{2}{tanx}}{\mathrm{1}+{tan}^{\mathrm{2}} {x}}\:=\frac{\mathrm{2}{t}}{\mathrm{1}+{t}^{\mathrm{2}} }\:\Rightarrow\int\frac{{dt}}{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)^{\mathrm{2}} }\:=\frac{\mathrm{1}}{\mathrm{2}}{arctan}\left({t}\right)\:+\frac{{t}}{\mathrm{2}\left(\mathrm{1}+{t}^{\mathrm{2}} \right)}\:+{c}_{\mathrm{2}} \\ $$$$\Rightarrow{I}\:=\frac{\mathrm{1}}{\mathrm{2}}{arctan}\left({t}\right)−\frac{{t}}{\mathrm{2}\left(\mathrm{1}+{t}^{\mathrm{2}} \right)}\:+{C} \\ $$

Pin

Leave a Reply

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