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Question Number 51991 by maxmathsup by imad last updated on 01/Jan/19
find f(a) =∫     (dx/( (√(1+ax^2 ))+(√(1−ax^2 ))))  with a>0
findf(a)=dx1+ax2+1ax2witha>0
Commented by Abdo msup. last updated on 05/Jan/19
changement x(√a)=t give  f(a)=∫        (dt/( (√a)((√(1+t^2 ))+(√(1−t^2 )))))  =(1/( (√a))) ∫     (dt/( (√(1+t^2 ))+(√(1−t^2 )))) ⇒(√a)f(a)=∫ (((√(1+t^2 ))−(√(1−t^2 )))/(2t^2 ))dt  =(1/2) ∫  ((√(1+t^2 ))/t^2 ) dt−(1/2) ∫  ((√(1−t^2 ))/t^2 ) dt  but by parts  ∫  ((√(1+t^2 ))/t^2 )dt =((−(√(1+t^2 )))/t) −∫ −(1/t)  ((2t)/(2(√(1+t^2 ))))dt  =−((√(1+t^2 ))/t) +ln(t+(√(1+t^2 ))) +c_1  also by psrts  ∫ ((√(1−t^2 ))/t^2 ) dt =−((√(1−t^2 ))/t) −∫  −(1/t) ((−2t)/(2(√(1−t^2 )))) dt  =−((√(1−t^2 ))/t) −arcsint  +c_2  ⇒  f(a)=(1/(2(√a))){ −((√(1+t^2 ))/t) +ln(t+(√(1+t^2 )))+((√(1−t^2 ))/t)  +arcsint } +C .
changementxa=tgivef(a)=dta(1+t2+1t2)=1adt1+t2+1t2af(a)=1+t21t22t2dt=121+t2t2dt121t2t2dtbutbyparts1+t2t2dt=1+t2t1t2t21+t2dt=1+t2t+ln(t+1+t2)+c1alsobypsrts1t2t2dt=1t2t1t2t21t2dt=1t2tarcsint+c2f(a)=12a{1+t2t+ln(t+1+t2)+1t2t+arcsint}+C.

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