Question

Evaluate : `int(x1+x^2)/(1+x^4)dx`

Answer 1

`I=int(x(1+x^2))/(1+x^4)dx`

`I=int(x(1+x^2)x  dx)/((1+x^4)-2x)`

Let `1+x^2 = t`
`2x  dx = dt`
`x  dx=1/2dt`

`I = 1/2int(txxdt)/(t^2 - 2(t-1))`

`I=1/2int(tdt)/(t^2-2t+2)`

`I=1/(2xx2)int(2t  dt)/(t^2-2t+2)`

`I = 1/4int((2t-2)+2 dt)/(t^2-2t+2)`

`I=1/4(int(2t -2)/(t^2 - 2t+2)dt+2int (dt)/(t^2-2t +2 ))`

`I = 1/4In|t^2-2t+2|+2/4 int (dt)/((t-1)^2 +1) + C_1`

`=1/4 In|t^2 - 2t+2|+1/2 tan^-1(t-1)+C_1`

`therefore 1 +x^2=t`
`=1/4In|(1+x^2)-2(1+x^2)+2|+1//2 tan^-1(1+x^2-1)+C`

`=1/4 In |1+x^4+2x^2 -2-2x^2|+1/2tan^-1 (x^2)+C`

`=1/4 In|x^4+1|+1/2 tan^-1(X^2)+C`