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Question Number 87103 by hamdhan last updated on 02/Apr/20

∫(dx/((1+x)(√(x−x^2 ))))

dx(1+x)xx2

Commented by abdomathmax last updated on 02/Apr/20

I =∫  (dx/((x+1)(√(−(x^2 −x))))) =∫  (dx/((x+1)(√(−(x^2 −x +(1/4)−(1/4))))))  =∫     (dx/((x+1)(√((1/4)−(x−(1/2))^2 )))) changement  x−(1/2)=(1/2)sint give  I =∫      ((cost dt)/(2((1/2)+(1/2)sint)(1/2)cost))  =∫   ((2dt)/(1+sint)) =_(tan((t/2))=u)     2 ∫   ((2du)/((1+u^2 )(1+((2u)/(1+u^2 )))))  =4 ∫    (du/(1+u^2  +2u)) =4 ∫  (du/((u+1)^2 )) =−(4/(1+u)) +C  =−(4/(1+tan((t/2)))) +C  t =arcsin(2x−1) ⇒  I  =((−4)/(1+tan((1/2)arcsin(2x−1)))) +C

I=dx(x+1)(x2x)=dx(x+1)(x2x+1414)=dx(x+1)14(x12)2changementx12=12sintgiveI=costdt2(12+12sint)12cost=2dt1+sint=tan(t2)=u22du(1+u2)(1+2u1+u2)=4du1+u2+2u=4du(u+1)2=41+u+C=41+tan(t2)+Ct=arcsin(2x1)I=41+tan(12arcsin(2x1))+C

Commented by Ar Brandon last updated on 02/Apr/20

Nice

Nice

Answered by hamdhan last updated on 02/Apr/20

x=sin^2 t;  dx=2sin tcos t dt  ∫((2sint.cost)/((1+sint)(√(sin^2 t−sin^4 t))))dt  ∫(2/((1+sint)))dt  ∫((2(1−sint))/(cos^2 t))dt  2∫(sec^2 t−tant.sect)dt  2tant−2sect + C  next step can do.....

x=sin2t;dx=2sintcostdt2sint.cost(1+sint)sin2tsin4tdt2(1+sint)dt2(1sint)cos2tdt2(sec2ttant.sect)dt2tant2sect+Cnextstepcando.....

Commented by Ar Brandon last updated on 04/Apr/20

I think you′ve mistakened something  somewhere.  It should be 1+sin^2 t instead of 1+sinx

Ithinkyouvemistakenedsomethingsomewhere.Itshouldbe1+sin2tinsteadof1+sinx

Answered by MJS last updated on 02/Apr/20

without trigonometric substitution:  ∫(dx/((1+x)(√(x−x^2 ))))=∫(dx/((1+x)(√(x(1−x)))))       [t=(√((1−x)/x)) → dx=−2(√(x^3 (1−x)))dt]  =−2∫(dt/(t^2 +2))=−(√2)arctan (t/(√3)) =  =−(√2)arctan (√((1−x)/(2x))) +C

withouttrigonometricsubstitution:dx(1+x)xx2=dx(1+x)x(1x)[t=1xxdx=2x3(1x)dt]=2dtt2+2=2arctant3==2arctan1x2x+C

Commented by john santu last updated on 03/Apr/20

super....sir

super....sir

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