Question Number 33145 by Ahmad Hajjaj last updated on 11/Apr/18
$$\underset{\:\mathrm{0}} {\overset{\mathrm{3}} {\int}}\:\frac{{dx}}{\:\sqrt{{x}+\mathrm{1}}\:+\:\sqrt{\mathrm{5}{x}+\mathrm{1}}}\:= \\ $$
Commented by prof Abdo imad last updated on 11/Apr/18
![I = ∫_0 ^3 (dx/( (√(x+1)) +(√(5x+1)))) .ch. (√(x+1)) =t ⇒ I = ∫_1 ^2 ((2tdt)/(t +(√(5(t^2 −1)+1)))) = ∫_1 ^2 ((2tdt)/(t +(√( 5t^2 −4)))) = ∫_1 ^2 ((2dt)/(1 +(√((5t^2 −4)/t^2 )))) = ∫_1 ^2 ((2dt)/(1+(√(5 −(4/t^2 ))))) after we use the ch. (2/t) =(√5) sinx ⇒ (t/2) = (1/( (√5) sinx)) ⇒t = (2/( (√5) sinx)) dt =−(2/( (√5))) ((cosx)/(sin^2 x)) and x =arcsin((2/(t(√5)))) I = ∫_(arcsin((2/( (√5))))) ^(arcsin((1/( (√5))))) (1/(1+(√5) (√(1−sin^2 x)))) −(2/( (√5))) ((cosx)/(sin^2 x))dx = −(2/( (√5)))∫_(arcsin((2/( (√5))))) ^(arcsin((1/( (√5))))) ((cosx)/(sin^2 x(1+(√5) cosx)))dx ∫_α ^β −((cosx)/(sin^2 x(1+(√5) cosx))) (by parts) = [ (1/(sinx)) (1/(1+(√5) cosx))]_α ^β − ∫_α ^β (1/(sinx)) (((√5) sinx)/((1+(√5) cosx)^2 ))dx =(1/(sinβ(1+(√5) cosβ))) − (1/(sinα(1+(√5) cosα))) −(√5) ∫_α ^β (dx/((1+(√5) cosx)^2 )) and this integral is calculable by ch. tan((x/2)) =u .... α=arcsin((2/( (√5)))) and β = arcsin((1/( (√5)))).](https://www.tinkutara.com/question/Q33149.png)
$${I}\:=\:\int_{\mathrm{0}} ^{\mathrm{3}} \:\:\:\:\:\:\:\:\frac{{dx}}{\:\sqrt{{x}+\mathrm{1}}\:\:+\sqrt{\mathrm{5}{x}+\mathrm{1}}}\:\:\:\:.{ch}.\:\sqrt{{x}+\mathrm{1}}\:={t}\:\Rightarrow \\ $$$${I}\:=\:\int_{\mathrm{1}} ^{\mathrm{2}} \:\:\:\:\:\:\:\:\frac{\mathrm{2}{tdt}}{{t}\:\:+\sqrt{\mathrm{5}\left({t}^{\mathrm{2}} −\mathrm{1}\right)+\mathrm{1}}} \\ $$$$=\:\int_{\mathrm{1}} ^{\mathrm{2}} \:\:\:\:\:\:\:\frac{\mathrm{2}{tdt}}{{t}\:+\sqrt{\:\mathrm{5}{t}^{\mathrm{2}} \:−\mathrm{4}}}\:\:=\:\int_{\mathrm{1}} ^{\mathrm{2}} \:\:\:\:\:\:\frac{\mathrm{2}{dt}}{\mathrm{1}\:+\sqrt{\frac{\mathrm{5}{t}^{\mathrm{2}} \:−\mathrm{4}}{{t}^{\mathrm{2}} }}} \\ $$$$=\:\int_{\mathrm{1}} ^{\mathrm{2}} \:\:\:\:\:\:\:\:\frac{\mathrm{2}{dt}}{\mathrm{1}+\sqrt{\mathrm{5}\:−\frac{\mathrm{4}}{{t}^{\mathrm{2}} }}}\:\:{after}\:{we}\:{use}\:{the}\:{ch}. \\ $$$$\frac{\mathrm{2}}{{t}}\:=\sqrt{\mathrm{5}}\:{sinx}\:\Rightarrow\:\frac{{t}}{\mathrm{2}}\:=\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}\:{sinx}}\:\Rightarrow{t}\:=\:\frac{\mathrm{2}}{\:\sqrt{\mathrm{5}}\:{sinx}} \\ $$$${dt}\:=−\frac{\mathrm{2}}{\:\sqrt{\mathrm{5}}}\:\:\frac{{cosx}}{{sin}^{\mathrm{2}} {x}}\:\:{and}\:{x}\:={arcsin}\left(\frac{\mathrm{2}}{{t}\sqrt{\mathrm{5}}}\right) \\ $$$${I}\:=\:\int_{{arcsin}\left(\frac{\mathrm{2}}{\:\sqrt{\mathrm{5}}}\right)} ^{{arcsin}\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}}\right)} \:\:\:\:\:\:\:\:\:\frac{\mathrm{1}}{\mathrm{1}+\sqrt{\mathrm{5}}\:\sqrt{\mathrm{1}−{sin}^{\mathrm{2}} {x}}}\:\:−\frac{\mathrm{2}}{\:\sqrt{\mathrm{5}}}\:\frac{{cosx}}{{sin}^{\mathrm{2}} {x}}{dx} \\ $$$$=\:−\frac{\mathrm{2}}{\:\sqrt{\mathrm{5}}}\int_{{arcsin}\left(\frac{\mathrm{2}}{\:\sqrt{\mathrm{5}}}\right)} ^{{arcsin}\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}}\right)} \:\:\:\:\:\:\:\frac{{cosx}}{{sin}^{\mathrm{2}} {x}\left(\mathrm{1}+\sqrt{\mathrm{5}}\:{cosx}\right)}{dx} \\ $$$$\int_{\alpha} ^{\beta} \:\:\:−\frac{{cosx}}{{sin}^{\mathrm{2}} {x}\left(\mathrm{1}+\sqrt{\mathrm{5}}\:{cosx}\right)}\:\left({by}\:{parts}\right) \\ $$$$=\:\left[\:\frac{\mathrm{1}}{{sinx}}\:\:\frac{\mathrm{1}}{\mathrm{1}+\sqrt{\mathrm{5}}\:{cosx}}\right]_{\alpha} ^{\beta} \:\:−\:\int_{\alpha} ^{\beta} \:\:\frac{\mathrm{1}}{{sinx}}\:\frac{\sqrt{\mathrm{5}}\:{sinx}}{\left(\mathrm{1}+\sqrt{\mathrm{5}}\:{cosx}\right)^{\mathrm{2}} }{dx} \\ $$$$=\frac{\mathrm{1}}{{sin}\beta\left(\mathrm{1}+\sqrt{\mathrm{5}}\:{cos}\beta\right)}\:−\:\frac{\mathrm{1}}{{sin}\alpha\left(\mathrm{1}+\sqrt{\mathrm{5}}\:{cos}\alpha\right)} \\ $$$$−\sqrt{\mathrm{5}}\:\:\int_{\alpha} ^{\beta} \:\:\:\:\frac{{dx}}{\left(\mathrm{1}+\sqrt{\mathrm{5}}\:{cosx}\right)^{\mathrm{2}} }\:\:{and}\:{this}\:{integral}\:{is} \\ $$$${calculable}\:{by}\:{ch}.\:{tan}\left(\frac{{x}}{\mathrm{2}}\right)\:={u}\:\:…. \\ $$$$\alpha={arcsin}\left(\frac{\mathrm{2}}{\:\sqrt{\mathrm{5}}}\right)\:{and}\:\beta\:=\:{arcsin}\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}}\right). \\ $$$$ \\ $$
Answered by MJS last updated on 11/Apr/18
![(1/(a+b))=((a−b)/((a+b)(a−b)))=((a−b)/(a^2 −b^2 )) (1/( (√(x+1))+(√(5x+1))))=(((√(x+1))−(√(5x+1)))/(x+1−5x−1))= =(((√(5x+1))−(√(x+1)))/(4x)) ∫_0 ^3 (dx/( (√(x+1))+(√(5x+1))))=(1/4)[(∫((√(5x+1))/x)dx−∫((√(x+1))/x)dx)]_0 ^3 = (∫((√(ax+b))/x)dx with b>0)= =2(√(ax+b))+(√b)×ln∣(((√(ax+b))−(√b))/( (√(ax+b))+(√b)))∣ =(1/4)[((2(√(5x+1))+ln∣(((√(5x+1))−1)/( (√(5x+1))+1))∣)−(2(√(x+1))+ln∣(((√(x+1))−1)/( (√(x+1))+1))∣))]_0 ^3 = =(1/4)[2((√(5x+1))−(√(x+1)))+ln((√(5x+1))−1)−ln((√(5x+1))+1)−ln((√(x+1))−1)+ln((√(x+1))+1)]_0 ^3 = =(1/4)[2((√(5x+1))−(√(x+1)))+ln(((√(x+1))+1)/( (√(5x+1))+1))+ln(((√(5x+1))−1)/( (√(x+1))−1))]_0 ^3 = lim_(x→0) (((√(5x+1))−1)/( (√(x+1))−1))=5 =1+(1/2)ln (3/5)](https://www.tinkutara.com/question/Q33150.png)
$$\frac{\mathrm{1}}{{a}+{b}}=\frac{{a}−{b}}{\left({a}+{b}\right)\left({a}−{b}\right)}=\frac{{a}−{b}}{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} } \\ $$$$ \\ $$$$\frac{\mathrm{1}}{\:\sqrt{{x}+\mathrm{1}}+\sqrt{\mathrm{5}{x}+\mathrm{1}}}=\frac{\sqrt{{x}+\mathrm{1}}−\sqrt{\mathrm{5}{x}+\mathrm{1}}}{{x}+\mathrm{1}−\mathrm{5}{x}−\mathrm{1}}= \\ $$$$=\frac{\sqrt{\mathrm{5}{x}+\mathrm{1}}−\sqrt{{x}+\mathrm{1}}}{\mathrm{4}{x}} \\ $$$$\underset{\mathrm{0}} {\overset{\mathrm{3}} {\int}}\frac{{dx}}{\:\sqrt{{x}+\mathrm{1}}+\sqrt{\mathrm{5}{x}+\mathrm{1}}}=\frac{\mathrm{1}}{\mathrm{4}}\left[\left(\int\frac{\sqrt{\mathrm{5}{x}+\mathrm{1}}}{{x}}{dx}−\int\frac{\sqrt{{x}+\mathrm{1}}}{{x}}{dx}\right)\right]_{\mathrm{0}} ^{\mathrm{3}} = \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\left(\int\frac{\sqrt{{ax}+{b}}}{{x}}{dx}\:\mathrm{with}\:{b}>\mathrm{0}\right)= \\ $$$$\:\:\:\:\:\:\:\:\:\:=\mathrm{2}\sqrt{{ax}+{b}}+\sqrt{{b}}×\mathrm{ln}\mid\frac{\sqrt{{ax}+{b}}−\sqrt{{b}}}{\:\sqrt{{ax}+{b}}+\sqrt{{b}}}\mid \\ $$$$ \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\left[\left(\left(\mathrm{2}\sqrt{\mathrm{5}{x}+\mathrm{1}}+\mathrm{ln}\mid\frac{\sqrt{\mathrm{5}{x}+\mathrm{1}}−\mathrm{1}}{\:\sqrt{\mathrm{5}{x}+\mathrm{1}}+\mathrm{1}}\mid\right)−\left(\mathrm{2}\sqrt{{x}+\mathrm{1}}+\mathrm{ln}\mid\frac{\sqrt{{x}+\mathrm{1}}−\mathrm{1}}{\:\sqrt{{x}+\mathrm{1}}+\mathrm{1}}\mid\right)\right)\right]_{\mathrm{0}} ^{\mathrm{3}} = \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\left[\mathrm{2}\left(\sqrt{\mathrm{5}{x}+\mathrm{1}}−\sqrt{{x}+\mathrm{1}}\right)+\mathrm{ln}\left(\sqrt{\mathrm{5}{x}+\mathrm{1}}−\mathrm{1}\right)−\mathrm{ln}\left(\sqrt{\mathrm{5}{x}+\mathrm{1}}+\mathrm{1}\right)−\mathrm{ln}\left(\sqrt{{x}+\mathrm{1}}−\mathrm{1}\right)+\mathrm{ln}\left(\sqrt{{x}+\mathrm{1}}+\mathrm{1}\right)\right]_{\mathrm{0}} ^{\mathrm{3}} = \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\left[\mathrm{2}\left(\sqrt{\mathrm{5}{x}+\mathrm{1}}−\sqrt{{x}+\mathrm{1}}\right)+\mathrm{ln}\frac{\sqrt{{x}+\mathrm{1}}+\mathrm{1}}{\:\sqrt{\mathrm{5}{x}+\mathrm{1}}+\mathrm{1}}+\mathrm{ln}\frac{\sqrt{\mathrm{5}{x}+\mathrm{1}}−\mathrm{1}}{\:\sqrt{{x}+\mathrm{1}}−\mathrm{1}}\right]_{\mathrm{0}} ^{\mathrm{3}} = \\ $$$$\:\:\:\:\:\:\:\:\:\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\sqrt{\mathrm{5}{x}+\mathrm{1}}−\mathrm{1}}{\:\sqrt{{x}+\mathrm{1}}−\mathrm{1}}=\mathrm{5} \\ $$$$=\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\:\frac{\mathrm{3}}{\mathrm{5}} \\ $$