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find-dx-x-x-1-x-2-x-




Question Number 122205 by mathmax by abdo last updated on 14/Nov/20
find ∫  (dx/(x(x+1)(√(x^2 +x))))
finddxx(x+1)x2+x
Commented by liberty last updated on 15/Nov/20
 ∫ (dx/((x^2 +x)^(3/2) )) = ∫ (dx/(x^3 (1+x^(−1) )^(3/2) ))   = ∫ ((x^(−2) dx)/(x(1+x^(−1) )^(3/2) )) ; [ u^(2/3)  = 1+(1/x) ]  ⇒ [ (2/3)u^(−(1/3))  du = −x^(−2)  dx ]  J = ∫ (u^(2/3) −1)(−(2/3)u^(−(1/3))  du)((1/u))  J= −(2/3)∫ (u^(2/3) −1)(u^(−(4/3)) ) du  J=−(2/3)∫ (u^(−(2/3)) −u^(−(4/3)) ) du   J = −(2/3)[ 3u^(1/3) +3u^(−(1/3))  ] + c   J = −2(√((x+1)/x))−2(√(x/(x+1))) + c   J=  −2(((x+1+x)/( (√(x^2 +x)))))+ c = −((4x+2)/( (√(x^2 +x)))) + c
dx(x2+x)32=dxx3(1+x1)32=x2dxx(1+x1)32;[u23=1+1x][23u13du=x2dx]J=(u231)(23u13du)(1u)J=23(u231)(u43)duJ=23(u23u43)duJ=23[3u13+3u13]+cJ=2x+1x2xx+1+cJ=2(x+1+xx2+x)+c=4x+2x2+x+c
Commented by benjo_mathlover last updated on 15/Nov/20
let J(x)=−((4x+2)/( (√(x^2 +x))))   ((dJ(x))/dx) = − [ ((4(√(x^2 +x))−(4x+2)(((2x+1)/(2(√(x^2 +x))))))/(x^2 +x)) ]        = − [((4(√(x^2 +x))−(((2x+1)^2 )/( (√(x^2 +x)))))/(x^2 +x)) ]        = − [ ((4x^2 +4x−4x^2 −4x−1)/((x^2 +x)(√(x^2 +x)))) ]        = (1/(x(x+1)(√(x^2 +x))))  correct sir Liberty.
letJ(x)=4x+2x2+xdJ(x)dx=[4x2+x(4x+2)(2x+12x2+x)x2+x]=[4x2+x(2x+1)2x2+xx2+x]=[4x2+4x4x24x1(x2+x)x2+x]=1x(x+1)x2+xcorrectsirLiberty.
Answered by TANMAY PANACEA last updated on 14/Nov/20
∫(((x+1)−x)/(x(x+1)(√(x^2 +x))))dx  ∫(dx/(x(√(x^2 +x))))−∫(dx/((x+1)(√(x^2 +x)) ))  a=(1/x)→dx=((−1)/a^2 ) da     b=(1/(x+1))→dx=−(1/b^2 )db  ∫((−da)/(a^2 ×(1/a)×(√((1/a^2 )+(1/a)))))−∫((−db)/(b^2 ×(1/b)(√(((1/b)−1)^2 +((1/b)−1)))))  ∫((−da)/( (√(1+a^2 ))))−∫(db/(b×(√((1/b^2 )−(2/b)+1+(1/b)−1))))  ∫((−da)/( (√(1+a^2 ))))−∫(db/( (√(1−b))))  ∫((−da)/( (√(1+a^2 ))))+∫((d(1−b))/( (√(1−b))))  −ln(a+(√(1+a^2 )) )+(((1−b)^(1/2) )/(1/2))+C  −ln((1/x)+(√(1+(1/x^2 ))) +2(1−(1/(x+1)))^(1/2) +c
(x+1)xx(x+1)x2+xdxdxxx2+xdx(x+1)x2+xa=1xdx=1a2dab=1x+1dx=1b2dbdaa2×1a×1a2+1adbb2×1b(1b1)2+(1b1)da1+a2dbb×1b22b+1+1b1da1+a2db1bda1+a2+d(1b)1bln(a+1+a2)+(1b)1212+Cln(1x+1+1x2+2(11x+1)12+c

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