Question

Evaluate:  `int (x(1+x^2))/(1+x^4)dx`

Answer 1

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

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

Let 1+x² t
      2x dx = dt
      `x dx=1/2 dt`

     `I=1/2 int(txxdt)/(t^2- 2(t-1))`

`I= 1/2 int(t  dt)/(t^2 -2t +2)`

`I= 1/(2xx2)int(2t  dt)/(t^2-2t+2)`

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

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

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

`=1/4In|t^2-2t+2| + 1/2  tan^-1 (t-1)+c_1`

`therefore 1+X^2=t`

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

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

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