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dxx3+1 , dxx4+1 , tanθdθ, cotθdθYue Kwok Choydxx3+1=dxx+1(x2-x+1) =131x+1-x-2x2-x+1dx (Break integrand by Partial Fraction)=13dxx+1-162x-1-3x2-x+1dx=13dxx+1-162x-1-3x2-x+1dx=13dxx+1-162x-1x2-x+1dx+121x2-x+1dx=13d(x+1)x+1-16d(x2-x+1)x2-x+1+121x-122+342dx=13lnx+1-16lnx2-x+1+13tan-12x-13+CFor I=1x-122+342dx, let x-12=34tanθ, dx=34sec2θdθ∴ I=34sec2θdθ34sec2θ=23dθ=23θ+C=23tan-12x-13+Cdxx4+1Method 1I=dxx4+1=dx(x2+1)2-2x2=dx(x2+2x+1)(x2-2x+1)The partial fraction is clumsy:1(x2+2x+1)(x2-2x+1)=Ax+Bx2+2x+1+Cx+Dx2-2x+1We get:Ax+Bx2-2x+1+Cx+Dx2+2x+1=1We need to solve for A, B, C, D with 4 equations involving surds.To avoid this, please study the following:(x2+2x)x2-2x-x2-2xx2+2x=0xx+2x2-2x-xx-2x2+2x=0Cancel x in (a), we get:(b)x+2x2-2x-x-2x2+2x=0(c)x+21-x-21=22Adding (b) and (c), we get:(d)x+2x2-2x+1-x-2x2+2x+1=22(e) 1(x2+2x+1)(x2-2x+1)=122x+2x2+2x+1-x-2x2-2x+1∴ I=122x+2x2+2x+1dx-122x-2x2-2x+1dx=142d(x2+2x+1)x2+2x+1+14dxx2+2x+1-142d(x2-2x+1)x2-2x+1+14dxx2-2x+1=142lnx2+2x+1x2-2x+1+14dxx+122+122+14dxx-122+122=28lnx2+2x+1x2-2x+1+24tan-12x+1+24tan-12x-1+C=28lnx2+2x+1x2-2x+1+24tan-12x+1+tan-12x-1+C=28lnx2+2x+1x2-2x+1+24tan-12x+1+2x-11-2x+12x-1+C=28lnx2+2x+1x2-2x+1+24tan-12x1-x2+C ….(1)Method 2I=dxx4+1=12(x2+1)-(x2-1)dxx4+1=12x2+1x4+1dx-12x2-1x4+1dx=12I1-12I2For I1=x2+1x4+1dx=1+1x2x2+1x2dxLet u=x-1x, du=1+1x2dxAnd x2+1x2=x-1x2+2=u2+2∴ I1=duu2+2=22tan-1u2+C1=22tan-112x-1x+C1=22tan-1x2-12x+C1For I2=x2-1x4+1dx=1-1x2x2+1x2dxLet u=x+1x, du=1-1x2dxAnd x2+1x2=x+1x2-2=u2-2∴ I2=duu2-2=duu-2u+2=122duu-2-duu+2=24lnu-2u+2+C2=24lnx+1x-2x+1x+2+C2=24lnx2-2x+1x2+2x+1+C2∴ I=12I1-12I2=24tan-1x2-12x-28lnx2-2x+1x2+2x+1+C=24tan-1x2-12x+28lnx2+2x+1x2-2x+1+C ….(2)Method 3 (Best)I=dxx4+1=12(x2+1)-(x2-1)dxx4+1=12x2+1x4+1dx-12x2-1x4+1=121+1x2x2+1x2dx-121-1x2x2+1x2dx=12dx-1xx-1x2+2dx-12dx+1xx+1x2-2=122tan-112x-1x-142lnx+1x-2x+1x+2+C …(3)Method 4(Using Complex numbers)1x4+1=1x2-ix2+i=12i1x2-i-1x2+i=12i1x2-i2-1x2+i2I=dxx4+1=12idxx2-i2-dxx2+i2=14iilnx-ix-i-12iitan-1xi+C …(4)The method is simple, but the appearance of many complex numbers seems uncomfortable.3.The result of Method 1 and Method 2 above differ by a constant.tan-1x2-12x-tan-12x1-x2=tan-1x2-12x-2x1-x21+x2-12x2x1-x2=tan-1∞=π24.tanθdθ, cotθdθdxx4+1=24tan-1x2-12x+28lnx2+2x+1x2-2x+1+C(a)Put x=tanθ,dx=sec2θ2tanθdθ dxx4+1=sec2θtan2θ+12tanθdθ=12cotθdθ∴cotθdθ=22tan-1tanθ-12tanθ+24lntanθ+2tanθ+1tanθ-2tanθ+1+C(b)Put x=cotθ,dx=-csc2θ2cotθdθ dxx4+1=-csc2θcot2θ+12cotθdθ=-12tanθdθtanθdθ=-22tan-1cotθ-12cotθ-24lncotθ+2cotθ+1cotθ-2cotθ+1+C ................
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