Question Number 104302 by bobhans last updated on 20/Jul/20

Commented bybobhans last updated on 21/Jul/20

by parts   { ((u=arc tan ((1/(2x^2 ))) ⇒du =((−(1/x^3 ))/(1+(1/(4x^4 )))) dx)),((dv = dx ⇒v = x)) :}  I= x arc tan ((1/(2x^2 )))+∫ ((4x^2 )/(4x^4 +1)) dx  I= x arctan ((1/(2x^2 ))) +∫ (x/(2x^2 −2x+1))−(x/(2x^2 +2x+1)) dx

Commented byjohn santu last updated on 20/Jul/20

I= x tan^(−1) ((1/(2x^2 )))+∫(x/(2x^2 −2x+1))−(x/(2x^2 +2x+1))dx  let J= ∫((2x)/(4x^2 −4x+2))−((2x)/(4x^2 +4x+2))dx  J= ∫ ((2x)/((2x−1)^2 +1))−((2x)/((2x+1)^2 +1))dx  J=∫((2x−1+1)/((2x−1)^2 +1))−((2x+1−1)/((2x+1)^2 +1))dx  J=∫ [((2x−1)/((2x−1)^2 +1))+(1/((2x−1)^2 +1))−((2x+1)/((2x+1)^2 +1))+(1/((2x+1)^2 +1))]dx  J= (1/2)ln ∣((2x^2 −2x+1)/(2x^2 +2x+1))∣ +(1/2){tan^(−1) (2x−1)+tan^(−1) (2x−1)}  we conclude   I= x tan^(−1) ((1/(2x^2 )))+(1/2)ln ∣((2x^2 −2x+1)/(2x^2 +2x+1))∣+  (1/2){tan^(−1) (2x−1)+tan^(−1) (2x+1)} + C  (JS ⊛)

Answered by OlafThorendsen last updated on 20/Jul/20

arctanu+arctan(1/u) = (π/2)  I = ∫arctan((1/(2x^2 )))dx  I = (π/2)x−∫arctan(2x^2 )dx  I = (π/2)x−xarctan(2x^2 )+∫((4x^2 )/(1+4x^4 ))dx  ...

Commented byAr Brandon last updated on 20/Jul/20

Cool !!!  Mind if I continue...  J=∫((4x^2 )/(1+4x^4 ))dx=∫{((2x^2 +1)/(4x^4 +1))+((2x^2 −1)/(4x^4 +1))}dx    =∫{((2+(1/x^2 ))/(4x^2 +(1/x^2 )))+((2−(1/x^2 ))/(4x^2 +(1/x^2 )))}dx=∫{((2+(1/x^2 ))/((2x−(1/x))^2 +4))+((2−(1/x^2 ))/((2x+(1/x))^2 −4))}dx    =(1/2)Arctan[((2x^2 −1)/(2x))]−(1/2)Arctanh[((2x^2 +1)/(2x))]+C

Answered by Dwaipayan Shikari last updated on 20/Jul/20

∫tan^(−1) ((1/(2x^2 )))dx=∫tan^(−1) (2x+1)−tan^(−1) (2x−1)dx  (1/2)∫tan^(−1) u−(1/2)∫tan^(−1) p  {take 2x+1=t   2x−1=p  (1/2)u tan^(−1) u−(1/2)∫(u/(u^2 +1))−(1/2)p tan^(−1) p−p+(1/2)∫(p/(p^2 +1))  (1/2)(2x+1)tan^(−1) (2x+1)−(1/4)log((2x+1)^2 +1)−(1/2)(2x−1)tan^(−1) (2x−1)  +(1/4)log((2x−1)^2 +1)+C

Commented byAr Brandon last updated on 20/Jul/20

wow ! that was creative 😃

Commented byDwaipayan Shikari last updated on 20/Jul/20

Sir olaftheorendsen , gives me this idea. 😃

Answered by mathmax by abdo last updated on 21/Jul/20

A =∫  arctan((1/(2x^2 )))dx   ⇒A= ∫((π/2)−arctan(2x^2 ))dx  A =((πx)/2) −∫ arctan(2x^2 )dx  by parts u^′  =1 and v =arctan(2x^2 ) ⇒  ∫ arctan(2x^2 )dx =x arctan(2x^2 )−∫ x.((4x)/(1+4x^4 ))dx  =xarctan(2x^2 )−4 ∫  (x^2 /(1+4x^4 )) dx   we have  ∫ (x^2 /(4x^4  +1))dx =∫  (x^2 /(((√2)x)^4  +1)) =_((√2)x =t)     ∫   (t^2 /(2(t^4  +1))) (dt/(√2))  =(1/(2(√2)))∫  (t^2 /(t^4  +1))dt =(1/(2(√2)))∫  (1/(t^2  +(1/t^2 )))dt =(1/(4(√2))) ∫ ((1−(1/t^2 )+1+(1/t^2 ))/(t^2  +(1/t^2 ))) dt  =(1/(4(√2))) ∫  ((1−(1/t^2 ))/((t+(1/t))^2 −2))dt(→u =t+(1/t))+(1/(4(√2))) ∫  ((1+(1/t^2 ))/((t−(1/t))^2 +2))dt(→v =t−(1/t))  =(1/(4(√2))) ∫ (du/(u^2 −2)) +(1/(4(√2))) ∫ (dv/(v^2  +2))  we have  ∫  (du/(u^2 −2)) =(1/(2(√2)))∫((1/(u−(√2)))−(1/(u+(√2))))du =(1/(2(√2)))ln∣((u−(√2))/(u+(√2)))∣+c_1  =(1/(2(√2)))ln∣((t+(1/t)−(√2))/(t+(1/t)+(√2)))∣+c_1   =(1/(2(√2)))ln∣(((√2)x+(1/((√2)x))−(√2))/((√2)x+(1/((√2)x))+(√2)))∣ +c_1   ∫ (dv/(v^2  +2)) =_(v=(√2)z)    ∫  (((√2)dz)/(2(1+z^2 ))) =((√2)/2) arctanz +c_2 =((√2)/2) arctan((v/(√2))) +c_2   =((√2)/2)arctan((1/(√2)){(√2)x−(1/((√2)x))}) +c_2  ⇒  A =x arctan(2x^2 )−(1/(√2)){(1/(2(√2)))ln∣((2x^2 +1−2x)/(2x^2 +1+2x))∣ +((√2)/2) arctan((1/(√2))((√2)x−(1/((√2)x)))) +C  A =xarctan(2x^2 )−(1/4)ln∣((2x^2 −2x+1)/(2x^2  +2x+1))∣−(1/2) arctan{x−(1/(2x))} +C

Commented bymathmax by abdo last updated on 21/Jul/20

sorry A =((πx)/2)−∫ arctan(2x^2 )dx ⇒  A =((πx)/2) −x arctan(2x^2 )+(1/4)ln∣((2x^2 −2x+1)/(2x^2  +2x+1))∣+(1/2) arctan(x−(1/(2x))) +C