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Question Number 161575 by amin96 last updated on 19/Dec/21
$$\int_{\mathrm{0}} ^{\mathrm{9}} \frac{\sqrt{{x}}}{\mathrm{1}−{x}}{dx}=? \\ $$$$ \\ $$
Answered by mathmax by abdo last updated on 19/Dec/21
![I =∫_0 ^9 ((√x)/(1−x))dx changement (√x)=t give I =∫_0 ^3 (t/(1−t^2 ))(2t)dt =2∫_0 ^3 (t^2 /(1−t^2 ))dt =−2∫_0 ^3 (t^2 /(t^2 −1))dt =−2∫_0 ^3 ((t^2 −1+1)/(t^2 −1))dt =−2∫_0 ^3 (1+(1/(t^2 −1)))dt =−6−2∫_0 ^3 (dt/((t−1)(t+1)) =−6−∫_0 ^3 ((1/(t−1))−(1/(t+1)))dt =−6−[ln∣((t−1)/(t+1))∣]_0 ^3 =−6−{ln((2/4))}=−6−ln((1/2))=−6+ln2](Q161580.png)
$$\mathrm{I}\:=\int_{\mathrm{0}} ^{\mathrm{9}} \:\frac{\sqrt{\mathrm{x}}}{\mathrm{1}−\mathrm{x}}\mathrm{dx}\:\:\mathrm{changement}\:\sqrt{\mathrm{x}}=\mathrm{t}\:\mathrm{give} \\ $$$$\mathrm{I}\:=\int_{\mathrm{0}} ^{\mathrm{3}} \:\frac{\mathrm{t}}{\mathrm{1}−\mathrm{t}^{\mathrm{2}} }\left(\mathrm{2t}\right)\mathrm{dt}\:=\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{3}} \frac{\mathrm{t}^{\mathrm{2}} }{\mathrm{1}−\mathrm{t}^{\mathrm{2}} }\mathrm{dt} \\ $$$$=−\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{3}} \:\frac{\mathrm{t}^{\mathrm{2}} }{\mathrm{t}^{\mathrm{2}} −\mathrm{1}}\mathrm{dt}\:=−\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{3}} \frac{\mathrm{t}^{\mathrm{2}} −\mathrm{1}+\mathrm{1}}{\mathrm{t}^{\mathrm{2}} −\mathrm{1}}\mathrm{dt} \\ $$$$=−\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{3}} \left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{t}^{\mathrm{2}} −\mathrm{1}}\right)\mathrm{dt}\:=−\mathrm{6}−\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{3}} \:\frac{\mathrm{dt}}{\left(\mathrm{t}−\mathrm{1}\right)\left(\mathrm{t}+\mathrm{1}\right.} \\ $$$$=−\mathrm{6}−\int_{\mathrm{0}} ^{\mathrm{3}} \left(\frac{\mathrm{1}}{\mathrm{t}−\mathrm{1}}−\frac{\mathrm{1}}{\mathrm{t}+\mathrm{1}}\right)\mathrm{dt} \\ $$$$=−\mathrm{6}−\left[\mathrm{ln}\mid\frac{\mathrm{t}−\mathrm{1}}{\mathrm{t}+\mathrm{1}}\mid\right]_{\mathrm{0}} ^{\mathrm{3}} =−\mathrm{6}−\left\{\mathrm{ln}\left(\frac{\mathrm{2}}{\mathrm{4}}\right)\right\}=−\mathrm{6}−\mathrm{ln}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)=−\mathrm{6}+\mathrm{ln2} \\ $$
Commented by amin96 last updated on 19/Dec/21
![diverges integral ∫_0 ^9 f(x)dx x[0;9] f(x)=((√x)/(1−x^2 )) x∈{0;1;2....9} x=1 f(x) =(1/0) ∫_0 ^9 ((√x)/(1−x^2 ))dx=(1/2)∫_0 ^3 (1/(1−t^2 ))dt= =(1/2)∫_0 ^1 (1/(1−t^2 ))dt+(1/2)∫_1 ^3 (1/(1−t^2 ))dt=lim_(u→1^− ) (1/2)∫_0 ^u (1/(1−t^2 ))dt+lim_(x→1^+ ) (1/2)∫_u ^3 (1/(1−t^2 ))dt= =.......continue](Q161581.png)
$${diverges}\:{integral}\:\int_{\mathrm{0}} ^{\mathrm{9}} {f}\left({x}\right){dx}\:\:\:\:{x}\left[\mathrm{0};\mathrm{9}\right]\:\:\:{f}\left({x}\right)=\frac{\sqrt{{x}}}{\mathrm{1}−{x}^{\mathrm{2}} }\:\:\:{x}\in\left\{\mathrm{0};\mathrm{1};\mathrm{2}....\mathrm{9}\right\} \\ $$$${x}=\mathrm{1}\:\:\:{f}\left({x}\right)\:=\frac{\mathrm{1}}{\mathrm{0}}\:\:\:\int_{\mathrm{0}} ^{\mathrm{9}} \frac{\sqrt{{x}}}{\mathrm{1}−{x}^{\mathrm{2}} }{dx}=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{3}} \frac{\mathrm{1}}{\mathrm{1}−{t}^{\mathrm{2}} }{dt}= \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}}{\mathrm{1}−{t}^{\mathrm{2}} }{dt}+\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{1}} ^{\mathrm{3}} \frac{\mathrm{1}}{\mathrm{1}−{t}^{\mathrm{2}} }{dt}=\underset{{u}\rightarrow\mathrm{1}^{−} } {\mathrm{lim}}\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{{u}} \frac{\mathrm{1}}{\mathrm{1}−{t}^{\mathrm{2}} }{dt}+\underset{{x}\rightarrow\mathrm{1}^{+} } {\mathrm{lim}}\frac{\mathrm{1}}{\mathrm{2}}\int_{{u}} ^{\mathrm{3}} \frac{\mathrm{1}}{\mathrm{1}−{t}^{\mathrm{2}} }{dt}= \\ $$$$=.......{continue} \\ $$
Commented by peter frank last updated on 20/Dec/21
$$\mathrm{thank}\:\mathrm{you} \\ $$
Commented by mathmax by abdo last updated on 20/Dec/21
$$\mathrm{not}\:\mathrm{correct}... \\ $$