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Question
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 x2= 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`
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