Question Number 145183 by mathmax by abdo last updated on 03/Jul/21
$$\mathrm{find}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{dx}}{\left(\sqrt{\mathrm{x}}+\sqrt{\mathrm{x}+\mathrm{1}}\right)^{\mathrm{3}} } \\ $$
Commented by justtry last updated on 03/Jul/21
Answered by mindispower last updated on 03/Jul/21
$${x}={sh}^{\mathrm{2}} \left({t}\right) \\ $$$$\Leftrightarrow\int\frac{{sh}\left(\mathrm{2}{t}\right){dt}}{{e}^{\mathrm{3}{t}} } \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}{e}^{{t}} }−\frac{{e}^{−\mathrm{5}{t}} }{\mathrm{2}} \\ $$$$=−\frac{{e}^{−{t}} }{\mathrm{2}}+\frac{{e}^{−\mathrm{5}{t}} }{\mathrm{10}}+{c},{t}={argsh}\left(\sqrt{{x}}\right) \\ $$
Answered by mathmax by abdo last updated on 03/Jul/21
![Ψ=∫_0 ^1 (dx/(((√x)+(√(x+1)))^3 )) changement x=sh^2 t give sht=(√x) ⇒t=argsh((√x))=log((√x)+(√(1+x))) ⇒ Ψ=∫_0 ^(log(1+(√2))) ((2sht cht)/((sht+cht)^3 )) dt =∫_0 ^(log(1+(√2))) ((sh(2t))/e^(3t) )dt =∫_0 ^(log(1+(√2))) e^(−3t) .((e^(2t) −e^(−2t) )/2)dt =(1/2)∫_0 ^(log(1+(√2))) (e^(−t) −e^(−5t) )dt =(1/2)[−e^(−t) +(1/5)e^(−5t) ]_0 ^(log(1+(√2))) Ψ=(1/2)(−(1/(1+(√2)))+(1/(5(1+(√2))^5 )) +1−(1/5))](https://www.tinkutara.com/question/Q145216.png)
$$\Psi=\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{dx}}{\left(\sqrt{\mathrm{x}}+\sqrt{\mathrm{x}+\mathrm{1}}\right)^{\mathrm{3}} }\:\:\mathrm{changement}\:\:\mathrm{x}=\mathrm{sh}^{\mathrm{2}} \mathrm{t}\:\mathrm{give} \\ $$$$\mathrm{sht}=\sqrt{\mathrm{x}}\:\Rightarrow\mathrm{t}=\mathrm{argsh}\left(\sqrt{\mathrm{x}}\right)=\mathrm{log}\left(\sqrt{\mathrm{x}}+\sqrt{\mathrm{1}+\mathrm{x}}\right)\:\Rightarrow \\ $$$$\Psi=\int_{\mathrm{0}} ^{\mathrm{log}\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)} \:\:\frac{\mathrm{2sht}\:\mathrm{cht}}{\left(\mathrm{sht}+\mathrm{cht}\right)^{\mathrm{3}} }\:\mathrm{dt}\:=\int_{\mathrm{0}} ^{\mathrm{log}\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)} \:\frac{\mathrm{sh}\left(\mathrm{2t}\right)}{\mathrm{e}^{\mathrm{3t}} }\mathrm{dt} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{log}\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)} \:\mathrm{e}^{−\mathrm{3t}} .\frac{\mathrm{e}^{\mathrm{2t}} −\mathrm{e}^{−\mathrm{2t}} }{\mathrm{2}}\mathrm{dt}\:=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{log}\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)} \:\left(\mathrm{e}^{−\mathrm{t}} \:−\mathrm{e}^{−\mathrm{5t}} \right)\mathrm{dt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left[−\mathrm{e}^{−\mathrm{t}} +\frac{\mathrm{1}}{\mathrm{5}}\mathrm{e}^{−\mathrm{5t}} \right]_{\mathrm{0}} ^{\mathrm{log}\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)} \\ $$$$\Psi=\frac{\mathrm{1}}{\mathrm{2}}\left(−\frac{\mathrm{1}}{\mathrm{1}+\sqrt{\mathrm{2}}}+\frac{\mathrm{1}}{\mathrm{5}\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)^{\mathrm{5}} }\:+\mathrm{1}−\frac{\mathrm{1}}{\mathrm{5}}\right) \\ $$