Question-201553

Question Number 201553 by Calculusboy last updated on 08/Dec/23

Answered by som(math1967) last updated on 09/Dec/23

1. ∫((1+logx−1)/((1+logx)^2 ))dx =∫(dx/((1+logx))) −∫(dx/((1+logx)^2 )) =(1/(1+logx))∫dx−∫{(d/dx)×(1/(1+logx))∫dx}dx −∫(dx/((1+logx)^2 )) =(x/(1+logx)) +∫((xdx)/(x(1+logx)^2 ))dx−∫(dx/((1+logx)^2 )) =(x/(1+logx)) +C

$$\mathrm{1}.\:\int\frac{\mathrm{1}+{logx}−\mathrm{1}}{\left(\mathrm{1}+{logx}\right)^{\mathrm{2}} }{dx} \\ $$$$=\int\frac{{dx}}{\left(\mathrm{1}+{logx}\right)}\:−\int\frac{{dx}}{\left(\mathrm{1}+{logx}\right)^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{1}}{\mathrm{1}+{logx}}\int{dx}−\int\left\{\frac{{d}}{{dx}}×\frac{\mathrm{1}}{\mathrm{1}+{logx}}\int{dx}\right\}{dx} \\ $$$$\:\:\:\:\:\:−\int\frac{{dx}}{\left(\mathrm{1}+{logx}\right)^{\mathrm{2}} } \\ $$$$=\frac{{x}}{\mathrm{1}+{logx}}\:+\int\frac{{xdx}}{{x}\left(\mathrm{1}+{logx}\right)^{\mathrm{2}} }{dx}−\int\frac{{dx}}{\left(\mathrm{1}+{logx}\right)^{\mathrm{2}} } \\ $$$$=\frac{{x}}{\mathrm{1}+{logx}}\:+{C} \\ $$

Commented by Calculusboy last updated on 09/Dec/23

$$\boldsymbol{{thanks}} \\ $$

Answered by som(math1967) last updated on 09/Dec/23

2.∫(e^((mtan^(−1) x)) /((1+x^2 )(√(1+x^2 ))))dx tan^(−1) x=t⇒(dx/(1+x^2 ))=dt (√(1+x^2 ))=(√(1+tan^2 t))=sect ∫((e^(mt) dt)/(sect)) I=∫coste^(mt) dt =cost∫e^(mt) dt−∫{(d/dt)cost∫e^(mt) dt}dt =((e^(mt) cost)/m) +(1/m)∫sinte^(mt) dt =((e^(mt) cost)/m^2 ) +((sint)/m)∫e^(mt) dt −(1/m)∫{(d/dt)sint∫e^(mt) dt}dt =((e^(mt) cost)/m^2 ) +((e^(mt) sint)/m^2 ) −(1/m^2 )∫e^(mt) costdt I=((e^(mt) (cost+sint))/m^2 )−(I/m^2 ) +C ⇒((I(m^2 +1))/m^2 ) =((e^(mt) (cost+sint))/m^2 ) +C ∴ I=((e^(mt) (cost+sint))/(m^2 +1)) +C_1 where t=tan^(−1) x

$$\mathrm{2}.\int\frac{{e}^{\left({m}\mathrm{tan}^{−\mathrm{1}} {x}\right)} }{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}{dx} \\ $$$$\:\mathrm{tan}^{−\mathrm{1}} {x}={t}\Rightarrow\frac{{dx}}{\mathrm{1}+{x}^{\mathrm{2}} }={dt} \\ $$$$\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }=\sqrt{\mathrm{1}+{tan}^{\mathrm{2}} {t}}={sect} \\ $$$$\:\int\frac{{e}^{{mt}} {dt}}{{sect}} \\ $$$$\:{I}=\int{coste}^{{mt}} {dt} \\ $$$$={cost}\int{e}^{{mt}} {dt}−\int\left\{\frac{{d}}{{dt}}{cost}\int{e}^{{mt}} {dt}\right\}{dt} \\ $$$$=\frac{{e}^{{mt}} {cost}}{{m}}\:+\frac{\mathrm{1}}{{m}}\int{sinte}^{{mt}} {dt} \\ $$$$=\frac{{e}^{{mt}} {cost}}{{m}^{\mathrm{2}} }\:+\frac{{sint}}{{m}}\int{e}^{{mt}} {dt} \\ $$$$\:−\frac{\mathrm{1}}{{m}}\int\left\{\frac{{d}}{{dt}}{sint}\int{e}^{{mt}} {dt}\right\}{dt} \\ $$$$=\frac{{e}^{{mt}} {cost}}{{m}^{\mathrm{2}} }\:+\frac{{e}^{{mt}} {sint}}{{m}^{\mathrm{2}} }\:−\frac{\mathrm{1}}{{m}^{\mathrm{2}} }\int{e}^{{mt}} {costdt} \\ $$$$\:{I}=\frac{{e}^{{mt}} \left({cost}+{sint}\right)}{{m}^{\mathrm{2}} }−\frac{{I}}{{m}^{\mathrm{2}} }\:+{C} \\ $$$$\Rightarrow\frac{{I}\left({m}^{\mathrm{2}} +\mathrm{1}\right)}{{m}^{\mathrm{2}} }\:=\frac{{e}^{{mt}} \left({cost}+{sint}\right)}{{m}^{\mathrm{2}} }\:+{C} \\ $$$$\:\therefore\:{I}=\frac{{e}^{{mt}} \left({cost}+{sint}\right)}{{m}^{\mathrm{2}} +\mathrm{1}}\:+{C}_{\mathrm{1}} \\ $$$$\:{where}\:{t}=\mathrm{tan}^{−\mathrm{1}} {x} \\ $$

Commented by Calculusboy last updated on 09/Dec/23

$$\boldsymbol{{thanks}} \\ $$

Answered by BaliramKumar last updated on 09/Dec/23

1. put 1+logx = t ⇒ dx = xdt logx = t−1 x = e^(t−1) ∫ (((t−1))/t^2 )∙e^(t−1) dt (1/e)∫ e^t ∙((1/t)−(1/t^2 ))dt ⇒ ∫e^x [f(x)+f^( ′) (x)]= e^x ∙f(x) +C (1/e)∙e^t ∙(1/t) + C = (e^(t−1) /t) + C (x/((1+logx))) + C

$$\mathrm{1}.\:\:\mathrm{put}\:\:\mathrm{1}+\mathrm{logx}\:=\:\mathrm{t}\:\:\:\:\:\:\:\:\:\Rightarrow\:\mathrm{dx}\:=\:\mathrm{xdt} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{logx}\:=\:\mathrm{t}−\mathrm{1} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{x}\:=\:\mathrm{e}^{\mathrm{t}−\mathrm{1}} \\ $$$$\int\:\frac{\left(\mathrm{t}−\mathrm{1}\right)}{\mathrm{t}^{\mathrm{2}} }\centerdot\mathrm{e}^{\mathrm{t}−\mathrm{1}} \mathrm{dt} \\ $$$$\frac{\mathrm{1}}{\mathrm{e}}\int\:\mathrm{e}^{\mathrm{t}} \centerdot\left(\frac{\mathrm{1}}{\mathrm{t}}−\frac{\mathrm{1}}{\mathrm{t}^{\mathrm{2}} }\right)\mathrm{dt}\:\:\Rightarrow\:\int\mathrm{e}^{\mathrm{x}} \left[\mathrm{f}\left(\mathrm{x}\right)+\mathrm{f}^{\:'} \left(\mathrm{x}\right)\right]=\:\mathrm{e}^{\mathrm{x}} \centerdot\mathrm{f}\left(\mathrm{x}\right)\:+\mathrm{C} \\ $$$$\frac{\mathrm{1}}{\mathrm{e}}\centerdot\mathrm{e}^{\mathrm{t}} \centerdot\frac{\mathrm{1}}{\mathrm{t}}\:+\:\mathrm{C}\:=\:\frac{\mathrm{e}^{\mathrm{t}−\mathrm{1}} }{\mathrm{t}}\:+\:\mathrm{C} \\ $$$$\frac{\mathrm{x}}{\left(\mathrm{1}+\mathrm{logx}\right)}\:+\:\mathrm{C} \\ $$$$ \\ $$

Commented by Calculusboy last updated on 09/Dec/23

$$\boldsymbol{{thanks}} \\ $$

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