Question Number 26759 by abdo imad last updated on 29/Dec/17
$${find}\:\:{the}\:{value}\:{of}\:\:\:\int_{\mathrm{0}} ^{\:\propto} \:\:\frac{{dx}}{\left({x}+\mathrm{1}\right)\left({x}+\mathrm{2}\right)\left({x}+\mathrm{3}\right)}\:. \\ $$
Commented by abdo imad last updated on 30/Dec/17
![we use the changement x+2=t I= ∫_0 ^∝ (dx/((x+1)(x+2)(x+3)))= ∫_2 ^(∝ ) (dt/((t−1)t(t+1))) and after decomposition I = ∫_2 ^∝ ( (1/(2(t−1))) − (1/t) + (1/(2(t+1))))dt = (1/2) ∫_2 ^∝ ( (1/(t−1)) + (1/(t+1)) )dt − ∫_2 ^∝ (dt/t) = (1/2) ∫_2 ^∝ ((2t)/(t^2 −1))dt − ∫_2 ^∝ (dt/t)=[ (1/2)ln/t^2 −1/ −ln/t/ ]_2 ^∝ =[ ln(((√(/t^2 −1/))/(/t/)) ]_2 ^∝ = [ln(((√(t^2 −1))/t))]_2 ^∝ =−ln(((√3)/2)) I= ln(2) −(1/2) ln(3).](https://www.tinkutara.com/question/Q26876.png)
$${we}\:{use}\:{the}\:{changement}\:{x}+\mathrm{2}={t} \\ $$$${I}=\:\int_{\mathrm{0}} ^{\propto} \:\:\frac{{dx}}{\left({x}+\mathrm{1}\right)\left({x}+\mathrm{2}\right)\left({x}+\mathrm{3}\right)}=\:\int_{\mathrm{2}} ^{\propto\:} \:\:\frac{{dt}}{\left({t}−\mathrm{1}\right){t}\left({t}+\mathrm{1}\right)}\:{and}\:{after}\:{decomposition} \\ $$$${I}\:=\:\int_{\mathrm{2}} ^{\propto} \left(\:\:\frac{\mathrm{1}}{\mathrm{2}\left({t}−\mathrm{1}\right)}\:−\:\frac{\mathrm{1}}{{t}}\:+\:\frac{\mathrm{1}}{\mathrm{2}\left({t}+\mathrm{1}\right)}\right){dt} \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{2}}\:\int_{\mathrm{2}} ^{\propto} \left(\:\frac{\mathrm{1}}{{t}−\mathrm{1}}\:+\:\frac{\mathrm{1}}{{t}+\mathrm{1}}\:\right){dt}\:−\:\int_{\mathrm{2}} ^{\propto} \:\frac{{dt}}{{t}} \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{2}}\:\int_{\mathrm{2}} ^{\propto} \:\frac{\mathrm{2}{t}}{{t}^{\mathrm{2}} −\mathrm{1}}{dt}\:−\:\int_{\mathrm{2}} ^{\propto} \frac{{dt}}{{t}}=\left[\:\frac{\mathrm{1}}{\mathrm{2}}{ln}/{t}^{\mathrm{2}} −\mathrm{1}/\:−{ln}/{t}/\:\right]_{\mathrm{2}} ^{\propto} \\ $$$$=\left[\:{ln}\left(\frac{\sqrt{/{t}^{\mathrm{2}} −\mathrm{1}/}}{/{t}/}\:\right]_{\mathrm{2}} ^{\propto} =\:\left[{ln}\left(\frac{\sqrt{{t}^{\mathrm{2}} −\mathrm{1}}}{{t}}\right)\right]_{\mathrm{2}} ^{\propto} =−{ln}\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right)\right. \\ $$$${I}=\:{ln}\left(\mathrm{2}\right)\:−\frac{\mathrm{1}}{\mathrm{2}}\:{ln}\left(\mathrm{3}\right). \\ $$
Answered by ajfour last updated on 29/Dec/17
$$\frac{\mathrm{1}}{\left({x}+\mathrm{1}\right)\left({x}+\mathrm{2}\right)\left({x}+\mathrm{3}\right)}=\frac{\mathrm{1}}{{t}\left({t}^{\mathrm{2}} −\mathrm{1}\right)} \\ $$$${if}\:{x}+\mathrm{2}={t} \\ $$$${let}\:{t}=\mathrm{sec}\:^{\mathrm{2}} \theta\:\:\:\Rightarrow\:\:{dt}=\mathrm{2sec}\:^{\mathrm{2}} \theta\mathrm{tan}\:\theta \\ $$$$\int_{\mathrm{2}} ^{\:\:\alpha+\mathrm{2}} \frac{{dt}}{{t}\left({t}^{\mathrm{2}} −\mathrm{1}\right)}=\mathrm{2}\int_{\theta_{\mathrm{1}} } ^{\:\theta_{\mathrm{2}} } \mathrm{cot}\:\theta{d}\theta \\ $$$$=\mathrm{2ln}\:\mid\frac{\mathrm{sin}\:\theta_{\mathrm{2}} }{\mathrm{sin}\:\theta_{\mathrm{1}} }\mid=\mathrm{ln}\:\left(\frac{\mathrm{1}−\frac{\mathrm{1}}{\alpha+\mathrm{2}}}{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}}\right) \\ $$$$=\mathrm{ln}\:\left(\mathrm{2}−\frac{\mathrm{2}}{\alpha+\mathrm{2}}\right)\:. \\ $$