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0-9-x-1-x-dx-

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Question Number 161575 by amin96 last updated on 19/Dec/21
∫_0 ^9 ((√x)/(1−x))dx=?
∫09x1−xdx=?
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
I=∫09x1−xdxchangementx=tgiveI=∫03t1−t2(2t)dt=2∫03t21−t2dt=−2∫03t2t2−1dt=−2∫03t2−1+1t2−1dt=−2∫03(1+1t2−1)dt=−6−2∫03dt(t−1)(t+1=−6−∫03(1t−1−1t+1)dt=−6−[ln∣t−1t+1∣]03=−6−{ln(24)}=−6−ln(12)=−6+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
divergesintegral∫09f(x)dxx[0;9]f(x)=x1−x2x∈{0;1;2….9}x=1f(x)=10∫09x1−x2dx=12∫0311−t2dt==12∫0111−t2dt+12∫1311−t2dt=limu→1−12∫0u11−t2dt+limx→1+12∫u311−t2dt==…….continue
Commented by peter frank last updated on 20/Dec/21
thank you
thankyou
Commented by mathmax by abdo last updated on 20/Dec/21
not correct...
notcorrect…

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