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calculate-0-1-ln-1-x-1-x-2-dx-




Question Number 64065 by mathmax by abdo last updated on 12/Jul/19
calculate ∫_0 ^1  ((ln(1+x))/(1+x^2 ))dx
calculate01ln(1+x)1+x2dx
Commented by mathmax by abdo last updated on 13/Jul/19
let f(t) =∫_0 ^1  ((ln(1+tx))/(1+x^2 ))dx  with t>0 we have  f^′ (t) =∫_0 ^1   (x/((1+tx)(1+x^2 ))) dx   let decompose   F(x) =  (x/((tx+1)(x^2  +1))) ⇒F(x)=(a/(tx+1)) +((bx +c)/(x^2  +1))  a= lim_(x→−(1/t))   (tx+1)F(x)=((−1)/(t((1/t^2 )+1))) =((−t^2 )/(t(1+t^2 ))) =−(t/(t^2  +1))  lim_(x→+∞)  xF(x) =0 =(a/t) +b ⇒b =−(a/t) =(1/(t^2  +1)) ⇒  F(x)=−(t/((t^2  +1)(tx+1))) +(((x/(t^2  +1)) +c)/(x^2  +1))  F(0) =0 =(t/(t^2  +1)) +c ⇒c=−(t/(t^2  +1)) ⇒  F(x)=(t/((t^2  +1)(tx+1))) +(1/(t^2  +1)) ((x−t)/(x^2  +1)) ⇒  f^′ (t)=∫_0 ^1  F(x)dx =(t/(t^2  +1)) ∫_0 ^1  (dx/(tx+1)) +(1/(t^2  +1)) ∫_0 ^1 ((x−t)/(x^2  +1)) dx  =(t/(t^2  +1))(1/t)[ln(tx+1)]_0 ^1   +(1/(2(t^2  +1)))∫_0 ^1   ((2x)/(x^2  +1)) dx−(t/(t^2  +1)) ∫_0 ^1  (dx/(x^2  +1))  =((ln(t+1))/(t^2  +1)) +(1/(2(t^2  +1)))ln(2) −(t/(t^2  +1))(π/4) ⇒  f(t) =∫_0 ^t    ((ln(1+u))/(u^2  +1))du+((ln(2))/2) ∫_0 ^t   (du/(1+u^2 )) −(π/8) ∫_0 ^t   ((2u)/(1+u^2 ))du +c  =∫_0 ^t  ((ln(1+u))/(1+u^2 )) +((ln(2))/2) arctan(t)−(π/8)ln(1+t^2 )  +c =∫_0 ^1 ((ln(1+tx))/(1+x^2 ))dx  ...be continued...
letf(t)=01ln(1+tx)1+x2dxwitht>0wehavef(t)=01x(1+tx)(1+x2)dxletdecomposeF(x)=x(tx+1)(x2+1)F(x)=atx+1+bx+cx2+1a=limx1t(tx+1)F(x)=1t(1t2+1)=t2t(1+t2)=tt2+1limx+xF(x)=0=at+bb=at=1t2+1F(x)=t(t2+1)(tx+1)+xt2+1+cx2+1F(0)=0=tt2+1+cc=tt2+1F(x)=t(t2+1)(tx+1)+1t2+1xtx2+1f(t)=01F(x)dx=tt2+101dxtx+1+1t2+101xtx2+1dx=tt2+11t[ln(tx+1)]01+12(t2+1)012xx2+1dxtt2+101dxx2+1=ln(t+1)t2+1+12(t2+1)ln(2)tt2+1π4f(t)=0tln(1+u)u2+1du+ln(2)20tdu1+u2π80t2u1+u2du+c=0tln(1+u)1+u2+ln(2)2arctan(t)π8ln(1+t2)+c=01ln(1+tx)1+x2dxbecontinued
Commented by mathmax by abdo last updated on 13/Jul/19
f(1) =∫_0 ^1  ((ln(1+x))/(1+x^2 ))dx =∫_0 ^1  ((ln(1+u))/(1+u^2 ))du +(π/8)ln(2)−(π/8)ln(2) +c ⇒  c=0 and   ∫_0 ^1   ((ln(1+tx))/(1+x^2 ))dx = ∫_0 ^t    ((ln(1+x))/(1+x^2 ))dx +((ln(2))/2) arctan(t)−(π/8)ln(1+t^2 )  ...be continued...
f(1)=01ln(1+x)1+x2dx=01ln(1+u)1+u2du+π8ln(2)π8ln(2)+cc=0and01ln(1+tx)1+x2dx=0tln(1+x)1+x2dx+ln(2)2arctan(t)π8ln(1+t2)becontinued
Commented by mathmax by abdo last updated on 14/Jul/19
let try another way  changement  x=tanθ give  I =∫_0 ^1  ((ln(1+x))/(1+x^2 ))dx = ∫_0 ^(π/4)  ((ln(1+tanθ))/(1+tan^2 θ))(1+tan^2 θ)dθ  =∫_0 ^(π/4) ln(1+tanθ)dθ   chang θ =(π/4)−u ⇒  ∫_0 ^(π/4)  ln(1+tanθ)dθ =∫_(π/4) ^0   ln(1+((1−tanu)/(1+tanu)))(−du)  =∫_0 ^(π/4)  ln((2/(1+tanu)))du =(π/4)ln(2)−∫_0 ^(π/4)  ln(1+tanu)du   ⇒ I =(π/4)ln(2)−I ⇒2I=(π/4)ln(2) ⇒  I=(π/8)ln(2)
lettryanotherwaychangementx=tanθgiveI=01ln(1+x)1+x2dx=0π4ln(1+tanθ)1+tan2θ(1+tan2θ)dθ=0π4ln(1+tanθ)dθchangθ=π4u0π4ln(1+tanθ)dθ=π40ln(1+1tanu1+tanu)(du)=0π4ln(21+tanu)du=π4ln(2)0π4ln(1+tanu)duI=π4ln(2)I2I=π4ln(2)I=π8ln(2)

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