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

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#### Solution

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

consider` x^2/((x^2+2)(2x^2+1))`

Put x^{2}= t (For finding partial fractions only)

`t/((t+2)(2t+1))=A/(t+2)+B/(2t+1)`

t=A(2t+1)+B(t+2)

On Solving we get A=2/3, B=-1/3

`t/((t+2)(2t+1))=(2/3)/(t+2)+(-1/3)/(2t+1)`

` x^2/((x^2+2)(2x^2+1))=(2/3)/(t+2)+(-1/3)/(2t+1)`

`I=int[(2/3)/(t+2)+(-1/3)/(2t+1)]dx`

`=2/3int 1/(x^2+2) dx-1/3int 1/(2x^2+1)dx`

`=2/3int 1/(x^2+(sqrt2)^2)dx-1/6int1/(x^2+(1/sqrt2)^2)dx`

`=sqrt2/3 tan^-1 (x/sqrt2)-1/(3sqrt2)tan^-1 (sqrt2 x)+c`

Concept: Methods of Integration: Integration Using Partial Fractions

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