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

Explanation

`I=int(x^(2))/((x^(2)+2)(2x^(2)+1))dx`
Let`" "(x^(2))/((x^(2)+2)(2x^(2)+1))=(A)/(x^(2)+2)+(B)/(2x^(2)+1)`
Put `x^(2)=y`
`(y)/((y+2)(2y+1))+(A)/(y+2)+(B)/(2y+1)`
`y=A(2y+1)+B(y+2)`
Put `y+2 =0 rArr y=-2`
`-2 = a(-3)+0`
`therefore" "A=(2)/(3)`
Put `2y+1=0 rArr y=-(1)/(2)`
`-(1)/(2)=0+B(-(1)/(2)+2)`
`-(1)/(2)=B((3)/(2))`
`therefore" "B=-(1)/(3)`
`therefore" "(y)/((y+2)(2y+1))=(2)/(3(y+2))-(1)/(3(2y+1))`
Then `(x^(2))/((x^(2)+2)(2x^(2)+1))=(2)/(3(x^(2)+2))-(1)/(3(2x^(2)+1))`
`int(x^(2))/((x^(2)+2)(2x^(2)+1))dx=int(2)/(3(x^(2)+2))dx-int(dx)/(3(2x^(2)+1))`
`=(2)/(3)int(dx)/(x^(2)+(sqrt2)^(2))-(1)/(3)int(dx)/(2x^(2)+1)`
`=(2)/(3)xx(1)/(sqrt2)tan^(-1).(x)/(sqrt2)-(1)/(3xx2)int(dx)/(x^(2)+((1)/(sqrt2))^(2))`
`=(sqrt2)/(3)tan^(-1).(x)/(sqrt2)-(1)/(6)xx(1)/(1//sqrt2) tan^(-1).(x)/(1//sqrt2)+c`
`=(sqrt2)/(3)tan^(-1).(x)/(3sqrt2)tan^(-1)sqrt2 x+c`