Question Number 35990 by abdo mathsup 649 cc last updated on 26/May/18
$${calculate}\:\int_{\mathrm{2}} ^{\mathrm{5}} \:\:\:\:\frac{{xdx}}{\mathrm{2}{x}+\mathrm{1}\:+\sqrt{{x}−\mathrm{1}}} \\ $$
Commented by abdo mathsup 649 cc last updated on 27/May/18
![let put I = ∫_2 ^5 ((xdx)/(2x+1+(√(x−1)))) changement (√(x−1))=t give x =t^2 +1 and I = ∫_1 ^2 ((t^2 +1)/(2t^2 +2+1 +t)) (2t)dt = ∫_1 ^2 ((2t^3 +2t)/(2t^2 +t+3))dt = ∫_1 ^2 ((t( 2t^2 +t+3) −t^2 −3t +2t)/(2t^2 +t +3))dt = ∫_1 ^2 t dt −∫_1 ^2 ((t^2 +t)/(2t^2 +t +3))dt but ∫_1 ^2 t dt =[(t^2 /2)]_1 ^2 = 2−(1/2) =(3/2) ∫_1 ^2 ((t^2 +t)/(2t^2 +t +3))dt =(1/2) ∫_1 ^2 ((2t^2 +t+t +3−3)/(2t^2 +t +3))dt =(1/2) ∫_1 ^2 dt +(1/2) ∫_1 ^2 ((t−3)/(2t^2 +t+3))dt =(1/2) +(1/8)∫_1 ^2 ((4t −12)/(2t^2 +t +3))dt =(1/2) +(1/8) J J = ∫_1 ^2 ((4t+1−13)/(2t^2 +t+3))dt =[ln(2t^2 +t+3)]_1 ^2 −13 ∫_1 ^2 (dt/(2t^2 +t+3)) = ln(13)−ln(6) −((13)/2) ∫_1 ^2 (dt/(t^2 +(t/2) +(3/2))) but t^2 +(t/2) +(3/2) =t^2 +2(1/4)t +(1/(16)) +(3/2) −(1/(16)) =(t +(1/4))^2 + ((23)/(16)) chang. t+(1/4) =((√(23))/4) x give ∫_1 ^2 (dt/(t^2 +(t/2)+(3/2))) = ∫_((1/( (√(23))))(5)) ^(9/( (√(23)))) (1/(((23)/(16))(1+x^2 ))) ((√(23))/4) dx = ((16)/(23)) ((√(23))/4) ∫_(5/( (√(23)))) ^(9/( (√(23)))) (dx/(1+x^2 )) =(4/( (√(23)))){ arctan((9/( (√(2))))))−arctan((5/( (√(23)))))} I =(3/2) −(1/2) −(1/8)J =1 −(1/8){ln(13)−ln(6)−((13)/2)(4/( (√(23)))){arctan((9/( (√(23)))))−arctan((5/( (√(23)))))}o](https://www.tinkutara.com/question/Q36044.png)
$${let}\:{put}\:{I}\:=\:\int_{\mathrm{2}} ^{\mathrm{5}} \:\:\:\frac{{xdx}}{\mathrm{2}{x}+\mathrm{1}+\sqrt{{x}−\mathrm{1}}}\:{changement} \\ $$$$\sqrt{{x}−\mathrm{1}}={t}\:{give}\:{x}\:={t}^{\mathrm{2}} \:+\mathrm{1}\:{and}\: \\ $$$${I}\:\:=\:\int_{\mathrm{1}} ^{\mathrm{2}} \:\:\:\:\:\:\frac{{t}^{\mathrm{2}} +\mathrm{1}}{\mathrm{2}{t}^{\mathrm{2}} \:+\mathrm{2}+\mathrm{1}\:+{t}}\:\left(\mathrm{2}{t}\right){dt} \\ $$$$=\:\int_{\mathrm{1}} ^{\mathrm{2}} \:\:\:\:\frac{\mathrm{2}{t}^{\mathrm{3}} \:+\mathrm{2}{t}}{\mathrm{2}{t}^{\mathrm{2}} \:+{t}+\mathrm{3}}{dt}\: \\ $$$$=\:\int_{\mathrm{1}} ^{\mathrm{2}} \:\:\:\frac{{t}\left(\:\mathrm{2}{t}^{\mathrm{2}} \:+{t}+\mathrm{3}\right)\:−{t}^{\mathrm{2}} \:−\mathrm{3}{t}\:+\mathrm{2}{t}}{\mathrm{2}{t}^{\mathrm{2}} \:+{t}\:+\mathrm{3}}{dt} \\ $$$$=\:\int_{\mathrm{1}} ^{\mathrm{2}} \:\:{t}\:{dt}\:\:−\int_{\mathrm{1}} ^{\mathrm{2}} \:\:\:\:\frac{{t}^{\mathrm{2}} \:+{t}}{\mathrm{2}{t}^{\mathrm{2}} \:+{t}\:+\mathrm{3}}{dt}\:{but} \\ $$$$\int_{\mathrm{1}} ^{\mathrm{2}} \:{t}\:{dt}\:=\left[\frac{{t}^{\mathrm{2}} }{\mathrm{2}}\right]_{\mathrm{1}} ^{\mathrm{2}} \:\:=\:\mathrm{2}−\frac{\mathrm{1}}{\mathrm{2}}\:=\frac{\mathrm{3}}{\mathrm{2}} \\ $$$$\int_{\mathrm{1}} ^{\mathrm{2}} \:\:\:\frac{{t}^{\mathrm{2}} \:+{t}}{\mathrm{2}{t}^{\mathrm{2}} \:+{t}\:+\mathrm{3}}{dt}\:\:=\frac{\mathrm{1}}{\mathrm{2}}\:\int_{\mathrm{1}} ^{\mathrm{2}} \:\:\frac{\mathrm{2}{t}^{\mathrm{2}} \:+{t}+{t}\:+\mathrm{3}−\mathrm{3}}{\mathrm{2}{t}^{\mathrm{2}} \:+{t}\:+\mathrm{3}}{dt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\:\int_{\mathrm{1}} ^{\mathrm{2}} {dt}\:\:\:+\frac{\mathrm{1}}{\mathrm{2}}\:\int_{\mathrm{1}} ^{\mathrm{2}} \:\:\frac{{t}−\mathrm{3}}{\mathrm{2}{t}^{\mathrm{2}} \:+{t}+\mathrm{3}}{dt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\:\:+\frac{\mathrm{1}}{\mathrm{8}}\int_{\mathrm{1}} ^{\mathrm{2}} \:\:\frac{\mathrm{4}{t}\:−\mathrm{12}}{\mathrm{2}{t}^{\mathrm{2}} \:+{t}\:+\mathrm{3}}{dt}\:=\frac{\mathrm{1}}{\mathrm{2}}\:+\frac{\mathrm{1}}{\mathrm{8}}\:{J} \\ $$$${J}\:=\:\int_{\mathrm{1}} ^{\mathrm{2}} \:\:\:\frac{\mathrm{4}{t}+\mathrm{1}−\mathrm{13}}{\mathrm{2}{t}^{\mathrm{2}} \:+{t}+\mathrm{3}}{dt}\:=\left[{ln}\left(\mathrm{2}{t}^{\mathrm{2}} \:+{t}+\mathrm{3}\right)\right]_{\mathrm{1}} ^{\mathrm{2}} \\ $$$$−\mathrm{13}\:\int_{\mathrm{1}} ^{\mathrm{2}} \:\:\:\:\frac{{dt}}{\mathrm{2}{t}^{\mathrm{2}} \:+{t}+\mathrm{3}}\:=\:{ln}\left(\mathrm{13}\right)−{ln}\left(\mathrm{6}\right) \\ $$$$−\frac{\mathrm{13}}{\mathrm{2}}\:\int_{\mathrm{1}} ^{\mathrm{2}} \:\:\:\:\frac{{dt}}{{t}^{\mathrm{2}} \:\:+\frac{{t}}{\mathrm{2}}\:+\frac{\mathrm{3}}{\mathrm{2}}}\:{but} \\ $$$${t}^{\mathrm{2}} \:+\frac{{t}}{\mathrm{2}}\:+\frac{\mathrm{3}}{\mathrm{2}}\:={t}^{\mathrm{2}} \:+\mathrm{2}\frac{\mathrm{1}}{\mathrm{4}}{t}\:\:+\frac{\mathrm{1}}{\mathrm{16}}\:+\frac{\mathrm{3}}{\mathrm{2}}\:−\frac{\mathrm{1}}{\mathrm{16}} \\ $$$$=\left({t}\:+\frac{\mathrm{1}}{\mathrm{4}}\right)^{\mathrm{2}} \:\:\:+\:\frac{\mathrm{23}}{\mathrm{16}}\:\:{chang}.\:{t}+\frac{\mathrm{1}}{\mathrm{4}}\:=\frac{\sqrt{\mathrm{23}}}{\mathrm{4}}\:{x}\:{give} \\ $$$$\int_{\mathrm{1}} ^{\mathrm{2}} \:\:\:\:\:\:\frac{{dt}}{{t}^{\mathrm{2}} \:+\frac{{t}}{\mathrm{2}}+\frac{\mathrm{3}}{\mathrm{2}}}\:=\:\int_{\frac{\mathrm{1}}{\:\sqrt{\mathrm{23}}}\left(\mathrm{5}\right)} ^{\frac{\mathrm{9}}{\:\sqrt{\mathrm{23}}}} \:\:\:\frac{\mathrm{1}}{\frac{\mathrm{23}}{\mathrm{16}}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)}\:\:\frac{\sqrt{\mathrm{23}}}{\mathrm{4}}\:{dx} \\ $$$$=\:\frac{\mathrm{16}}{\mathrm{23}}\:\frac{\sqrt{\mathrm{23}}}{\mathrm{4}}\:\int_{\frac{\mathrm{5}}{\:\sqrt{\mathrm{23}}}} ^{\frac{\mathrm{9}}{\:\sqrt{\mathrm{23}}}} \:\:\:\:\frac{{dx}}{\mathrm{1}+{x}^{\mathrm{2}} }\:=\frac{\mathrm{4}}{\:\sqrt{\mathrm{23}}}\left\{\:{arctan}\left(\frac{\mathrm{9}}{\:\sqrt{\left.\mathrm{2}\right)}}\right)−{arctan}\left(\frac{\mathrm{5}}{\:\sqrt{\mathrm{23}}}\right)\right\} \\ $$$${I}\:=\frac{\mathrm{3}}{\mathrm{2}}\:−\frac{\mathrm{1}}{\mathrm{2}}\:−\frac{\mathrm{1}}{\mathrm{8}}{J} \\ $$$$=\mathrm{1}\:−\frac{\mathrm{1}}{\mathrm{8}}\left\{{ln}\left(\mathrm{13}\right)−{ln}\left(\mathrm{6}\right)−\frac{\mathrm{13}}{\mathrm{2}}\frac{\mathrm{4}}{\:\sqrt{\mathrm{23}}}\left\{{arctan}\left(\frac{\mathrm{9}}{\:\sqrt{\mathrm{23}}}\right)−{arctan}\left(\frac{\mathrm{5}}{\:\sqrt{\mathrm{23}}}\right)\right\}{o}\right. \\ $$
Commented by abdo mathsup 649 cc last updated on 27/May/18
$${I}\:=\:\mathrm{1}\:−\frac{{ln}\left(\mathrm{13}\right)}{\mathrm{8}}\:\:+\frac{{ln}\left(\mathrm{6}\right)}{\mathrm{8}}\:\:+\frac{\mathrm{13}}{\mathrm{4}\sqrt{\mathrm{23}}}\left\{\:{arctan}\left(\frac{\mathrm{9}}{\:\sqrt{\mathrm{23}}}\right)−{arctan}\left(\frac{\mathrm{5}}{\:\sqrt{\mathrm{23}}}\right)\right\}. \\ $$