Question Number 40132 by maxmathsup by imad last updated on 16/Jul/18
$${calculate}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\frac{{dt}}{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)^{\mathrm{3}} } \\ $$
Commented by maxmathsup by imad last updated on 21/Jul/18
![let I = ∫_0 ^1 (dt/((1+t^2 )^2 )) I = ∫_0 ^1 ((1+t^2 )/((1+t^2 )^3 ))dt = ∫_0 ^1 (dt/((1+t^2 )^3 )) +∫_0 ^1 (t^2 /((1+t^2 )^3 ))dt⇒ ∫_0 ^1 (dt/((1+t^2 )^3 )) = I −∫_0 ^1 (t^2 /((1+t^2 )^3 )) changement t=tanx give I = ∫_0 ^(π/4) (((1+tan^2 x)dx)/((1+tan^2 x)^2 )) = ∫_0 ^(π/4) (dx/(1+tan^2 x)) =∫_0 ^(π/4) cos^2 xdx =∫_0 ^(π/4) ((1+cos(2x))/2)dx =(π/8) +(1/4)[sin(2x)]_0 ^(π/4) =(π/8) +(1/4) also by parts ∫_0 ^1 t (1+t^2 )^(−3) t dt =−(1/4) ∫_0 ^1 −4t(1+t^2 )^(−3) t dt =−(1/4){[ (1+t^2 )^(−2) t]_0 ^1 −∫_0 ^1 (dt/((1+t^2 )^2 ))} =−(1/4) 2^(−2) +(1/4)((π/8) +(1/4)) = (π/(32)) +(1/(16)) −(1/(16)) =(π/(32)) ⇒ ∫_0 ^1 (dt/((1+t^2 )^3 )) =(π/8) +(1/4) −(π/(32)) =((3π)/(32)) +(1/4) .](https://www.tinkutara.com/question/Q40430.png)
$${let}\:{I}\:\:=\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\:\:\frac{{dt}}{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)^{\mathrm{2}} } \\ $$$${I}\:=\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{\mathrm{1}+{t}^{\mathrm{2}} }{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)^{\mathrm{3}} }{dt}\:=\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\frac{{dt}}{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)^{\mathrm{3}} }\:\:+\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\frac{{t}^{\mathrm{2}} }{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)^{\mathrm{3}} }{dt}\Rightarrow \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\:\frac{{dt}}{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)^{\mathrm{3}} }\:=\:{I}\:−\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\:\frac{{t}^{\mathrm{2}} }{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)^{\mathrm{3}} }\:\:{changement}\:{t}={tanx}\:{give} \\ $$$${I}\:=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:\:\frac{\left(\mathrm{1}+{tan}^{\mathrm{2}} {x}\right){dx}}{\left(\mathrm{1}+{tan}^{\mathrm{2}} {x}\right)^{\mathrm{2}} }\:=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:\:\frac{{dx}}{\mathrm{1}+{tan}^{\mathrm{2}} {x}}\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:{cos}^{\mathrm{2}} {xdx} \\ $$$$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:\:\frac{\mathrm{1}+{cos}\left(\mathrm{2}{x}\right)}{\mathrm{2}}{dx}\:=\frac{\pi}{\mathrm{8}}\:+\frac{\mathrm{1}}{\mathrm{4}}\left[{sin}\left(\mathrm{2}{x}\right)\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:=\frac{\pi}{\mathrm{8}}\:+\frac{\mathrm{1}}{\mathrm{4}}\:{also}\:{by}\:{parts} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} {t}\:\left(\mathrm{1}+{t}^{\mathrm{2}} \right)^{−\mathrm{3}} {t}\:{dt}\:\:=−\frac{\mathrm{1}}{\mathrm{4}}\:\int_{\mathrm{0}} ^{\mathrm{1}} −\mathrm{4}{t}\left(\mathrm{1}+{t}^{\mathrm{2}} \right)^{−\mathrm{3}} \:{t}\:{dt} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{4}}\left\{\left[\:\left(\mathrm{1}+{t}^{\mathrm{2}} \right)^{−\mathrm{2}} \:{t}\right]_{\mathrm{0}} ^{\mathrm{1}} \:\:−\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\:\frac{{dt}}{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)^{\mathrm{2}} }\right\} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{4}}\:\mathrm{2}^{−\mathrm{2}} \:\:+\frac{\mathrm{1}}{\mathrm{4}}\left(\frac{\pi}{\mathrm{8}}\:+\frac{\mathrm{1}}{\mathrm{4}}\right)\:=\:\frac{\pi}{\mathrm{32}}\:+\frac{\mathrm{1}}{\mathrm{16}}\:−\frac{\mathrm{1}}{\mathrm{16}}\:=\frac{\pi}{\mathrm{32}}\:\Rightarrow \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\:\frac{{dt}}{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)^{\mathrm{3}} }\:=\frac{\pi}{\mathrm{8}}\:+\frac{\mathrm{1}}{\mathrm{4}}\:−\frac{\pi}{\mathrm{32}}\:=\frac{\mathrm{3}\pi}{\mathrm{32}}\:+\frac{\mathrm{1}}{\mathrm{4}}\:. \\ $$$$ \\ $$