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solve-x-4-x-2-1-2-x-2-1-dx-




Question Number 187855 by Michaelfaraday last updated on 23/Feb/23
solve  ∫((x^4 +x^2 +1)/(2(x^2 +1)))dx
solvex4+x2+12(x2+1)dx
Answered by cortano12 last updated on 23/Feb/23
 I=(1/2)∫ (((x^2 +1)^2 −x^2 )/(x^2 +1)) dx   I=(1/6)x^3 +(1/2)x−(1/2)∫ (x^2 /(x^2 +1)) dx   = (1/6)x^3 +(1/2)x−(1/2)∫ ((tan^2 θ)/(sec^2 θ)) sec^2 θ dθ ; x=tan θ   = (1/6)x^3 +(1/2)x−(1/2)∫(sec^2 θ−1)dθ   = (1/6)x^3 +(1/2)x−(1/2)x+(1/2)arctan x+C   = (1/6)x^3 +(1/2)arctan x+C
I=12(x2+1)2x2x2+1dxI=16x3+12x12x2x2+1dx=16x3+12x12tan2θsec2θsec2θdθ;x=tanθ=16x3+12x12(sec2θ1)dθ=16x3+12x12x+12arctanx+C=16x3+12arctanx+C
Commented by Michaelfaraday last updated on 25/Feb/23
thanks sir
thankssir
Answered by CElcedricjunior last updated on 23/Feb/23
∫((x^4 +x^2 +1)/(2(x^2 +1)))dx=(1/2)∫[((x^2 (x^2 +1))/((x^2 +1)))+(1/((x^2 +1)))]dx  =(1/2)∫[x^2 +(1/(x^2 +1))]dx ■Moivre  =(1/2)[(1/3)x^3 +arctan(x)]+k ★Cedric junior
x4+x2+12(x2+1)dx=12[x2(x2+1)(x2+1)+1(x2+1)]dx=12[x2+1x2+1]dx◼Moivre=12[13x3+arctan(x)]+kCedricjunior
Commented by Michaelfaraday last updated on 25/Feb/23
thanks sir
thankssir

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