Question and Answers Forum

All Questions      Topic List

Integration Questions

Previous in All Question      Next in All Question      

Previous in Integration      Next in Integration      

Question Number 36397 by prof Abdo imad last updated on 01/Jun/18

find I = ∫_1 ^2 (dx/(x(√(x+1)) +(x+1)(√x)))

$${find}\:{I}\:\:=\:\int_{\mathrm{1}} ^{\mathrm{2}} \:\:\:\:\:\frac{{dx}}{{x}\sqrt{{x}+\mathrm{1}}\:\:+\left({x}+\mathrm{1}\right)\sqrt{{x}}} \\ $$$$ \\ $$

Answered by behi83417@gmail.com last updated on 01/Jun/18

x=tg^2 t⇒dx=2tgt(1+tg^2 t)dt (1/(cos^2 t))=1+x⇒cost=(1/(√(1+x))) I=∫((2tgt(1+tg^2 t)dt)/(tg^2 t.(1/(cost))+(1/(cos^2 t)).tgt))=∫((2tgt(1+tg^2 t)dt)/((sin^2 t+sint)/(cos^3 t)))= =∫((2sintdt)/(sin^2 t+sint))=∫((2dt)/(1+sint))=((−4)/(1+tg(t/2)))+c tg(t/2)=(√((1−cost)/(1+cost)))=(√((1−(1/(√(1+x))))/(1+(1/(√(1+x))))))= =(√(((√(1+x))−1)/((√(1+x))+1)))=(√((((√(1+x))−1)^2 )/x))=(((√(1+x))−1)/(√x)) ⇒I=((−4)/(1+(((√(1+x))−1)/(√x))))+c=((−4(√x))/((√x)+(√(1+x))−1))+c I=F(2)−F(1)=(4/(√2))−((4(√2))/((√2)+(√3)−1))= =2(√2)+4(√6)+10+2(√3)+2(√2)=2(2(√6)+(√3)+2(√2)+5).■ ((4(√2))/((√3)+(√2)−1))×(((√3)+(√2)+1)/((√3)+(√2)+1))=(((√3)+(√2)+1)/(2(2−(√6))))×((2+(√6))/(2+(√6)))×4(√2)= =−(√2)(2(√3)+2(√2)+2+3(√2)+2(√3)+(√6))= =−(√2)(4(√3)+5(√2)+(√6)+2)=−4(√6)−10−2(√3)−2(√2)

$${x}={tg}^{\mathrm{2}} {t}\Rightarrow{dx}=\mathrm{2}{tgt}\left(\mathrm{1}+{tg}^{\mathrm{2}} {t}\right){dt} \\ $$$$\frac{\mathrm{1}}{{cos}^{\mathrm{2}} {t}}=\mathrm{1}+{x}\Rightarrow{cost}=\frac{\mathrm{1}}{\sqrt{\mathrm{1}+{x}}} \\ $$$${I}=\int\frac{\mathrm{2}{tgt}\left(\mathrm{1}+{tg}^{\mathrm{2}} {t}\right){dt}}{{tg}^{\mathrm{2}} {t}.\frac{\mathrm{1}}{{cost}}+\frac{\mathrm{1}}{{cos}^{\mathrm{2}} {t}}.{tgt}}=\int\frac{\mathrm{2}{tgt}\left(\mathrm{1}+{tg}^{\mathrm{2}} {t}\right){dt}}{\frac{{sin}^{\mathrm{2}} {t}+{sint}}{{cos}^{\mathrm{3}} {t}}}= \\ $$$$=\int\frac{\mathrm{2}{sintdt}}{{sin}^{\mathrm{2}} {t}+{sint}}=\int\frac{\mathrm{2}{dt}}{\mathrm{1}+{sint}}=\frac{−\mathrm{4}}{\mathrm{1}+{tg}\frac{{t}}{\mathrm{2}}}+{c} \\ $$$${tg}\frac{{t}}{\mathrm{2}}=\sqrt{\frac{\mathrm{1}−{cost}}{\mathrm{1}+{cost}}}=\sqrt{\frac{\mathrm{1}−\frac{\mathrm{1}}{\sqrt{\mathrm{1}+{x}}}}{\mathrm{1}+\frac{\mathrm{1}}{\sqrt{\mathrm{1}+{x}}}}}= \\ $$$$=\sqrt{\frac{\sqrt{\mathrm{1}+{x}}−\mathrm{1}}{\sqrt{\mathrm{1}+{x}}+\mathrm{1}}}=\sqrt{\frac{\left(\sqrt{\mathrm{1}+{x}}−\mathrm{1}\right)^{\mathrm{2}} }{{x}}}=\frac{\sqrt{\mathrm{1}+{x}}−\mathrm{1}}{\sqrt{{x}}} \\ $$$$\Rightarrow{I}=\frac{−\mathrm{4}}{\mathrm{1}+\frac{\sqrt{\mathrm{1}+{x}}−\mathrm{1}}{\sqrt{{x}}}}+{c}=\frac{−\mathrm{4}\sqrt{{x}}}{\sqrt{{x}}+\sqrt{\mathrm{1}+{x}}−\mathrm{1}}+{c} \\ $$$${I}={F}\left(\mathrm{2}\right)−{F}\left(\mathrm{1}\right)=\frac{\mathrm{4}}{\sqrt{\mathrm{2}}}−\frac{\mathrm{4}\sqrt{\mathrm{2}}}{\sqrt{\mathrm{2}}+\sqrt{\mathrm{3}}−\mathrm{1}}= \\ $$$$=\mathrm{2}\sqrt{\mathrm{2}}+\mathrm{4}\sqrt{\mathrm{6}}+\mathrm{10}+\mathrm{2}\sqrt{\mathrm{3}}+\mathrm{2}\sqrt{\mathrm{2}}=\mathrm{2}\left(\mathrm{2}\sqrt{\mathrm{6}}+\sqrt{\mathrm{3}}+\mathrm{2}\sqrt{\mathrm{2}}+\mathrm{5}\right).\blacksquare \\ $$$$\frac{\mathrm{4}\sqrt{\mathrm{2}}}{\sqrt{\mathrm{3}}+\sqrt{\mathrm{2}}−\mathrm{1}}×\frac{\sqrt{\mathrm{3}}+\sqrt{\mathrm{2}}+\mathrm{1}}{\sqrt{\mathrm{3}}+\sqrt{\mathrm{2}}+\mathrm{1}}=\frac{\sqrt{\mathrm{3}}+\sqrt{\mathrm{2}}+\mathrm{1}}{\mathrm{2}\left(\mathrm{2}−\sqrt{\mathrm{6}}\right)}×\frac{\mathrm{2}+\sqrt{\mathrm{6}}}{\mathrm{2}+\sqrt{\mathrm{6}}}×\mathrm{4}\sqrt{\mathrm{2}}= \\ $$$$=−\sqrt{\mathrm{2}}\left(\mathrm{2}\sqrt{\mathrm{3}}+\mathrm{2}\sqrt{\mathrm{2}}+\mathrm{2}+\mathrm{3}\sqrt{\mathrm{2}}+\mathrm{2}\sqrt{\mathrm{3}}+\sqrt{\mathrm{6}}\right)= \\ $$$$=−\sqrt{\mathrm{2}}\left(\mathrm{4}\sqrt{\mathrm{3}}+\mathrm{5}\sqrt{\mathrm{2}}+\sqrt{\mathrm{6}}+\mathrm{2}\right)=−\mathrm{4}\sqrt{\mathrm{6}}−\mathrm{10}−\mathrm{2}\sqrt{\mathrm{3}}−\mathrm{2}\sqrt{\mathrm{2}} \\ $$$$ \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 01/Jun/18

∫_1 ^2 (dx/((√x) ×(√(x+1)) ((√(x+1)) +(√x) ))) ∫_1 ^2 (((√(x+1_( ) )) −(√x))/((√(x+1)) ×(√x)))dx ∫_1 ^2 (dx/((√x) ))−∫_1 ^2 (dx/(√(x+1))) =∣(((√x) )/(1/2))−(((√(x+1)) )/(1/2))∣_1 ^2 =2{((√2) −(√(2+1)))−((√1) −(√2) } 2((√2) −(√(3))) −((√1) −(√2) )} 2(−(√3) +2(√2) −1)

$$\int_{\mathrm{1}} ^{\mathrm{2}} \frac{{dx}}{\sqrt{{x}}\:×\sqrt{{x}+\mathrm{1}}\:\left(\sqrt{{x}+\mathrm{1}}\:+\sqrt{{x}}\:\right)} \\ $$$$\int_{\mathrm{1}} ^{\mathrm{2}} \frac{\sqrt{{x}+\mathrm{1}_{\:\:} }\:−\sqrt{{x}}}{\sqrt{{x}+\mathrm{1}}\:×\sqrt{{x}}}{dx} \\ $$$$\int_{\mathrm{1}} ^{\mathrm{2}} \frac{{dx}}{\sqrt{{x}}\:}−\int_{\mathrm{1}} ^{\mathrm{2}} \frac{{dx}}{\sqrt{{x}+\mathrm{1}}}\: \\ $$$$=\mid\frac{\sqrt{{x}}\:}{\frac{\mathrm{1}}{\mathrm{2}}}−\frac{\sqrt{{x}+\mathrm{1}}\:}{\frac{\mathrm{1}}{\mathrm{2}}}\mid_{\mathrm{1}} ^{\mathrm{2}} \\ $$$$=\mathrm{2}\left\{\left(\sqrt{\mathrm{2}}\:−\sqrt{\mathrm{2}+\mathrm{1}}\right)−\left(\sqrt{\mathrm{1}}\:−\sqrt{\mathrm{2}}\:\right\}\right. \\ $$$$\mathrm{2}\left(\sqrt{\mathrm{2}}\:−\sqrt{\left.\mathrm{3}\right)}\:−\left(\sqrt{\mathrm{1}}\:−\sqrt{\mathrm{2}}\:\right)\right\} \\ $$$$\mathrm{2}\left(−\sqrt{\mathrm{3}}\:\:+\mathrm{2}\sqrt{\mathrm{2}}\:−\mathrm{1}\right) \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com