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प्रश्न
Evaluate:
`int x^2/(x^4+x^2-2)dx`
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उत्तर
`int x^2/(x^4+x^2-2)dx`
`=int x^2/((x^2-1)(x^2+2))dx`
`=int x^2/((x+1)(x-1)(x^2+2))dx`
Using partial fraction
`x^/((x+1)(x-1)(x^2+2))=A/(x-1)+B/(x+1)+(Cx+D)/(x^2+2)`
`=(A(x+1)(x^2+2)+B(x-1)(x^2+2)+(Cx+D)(x+1)(x-1))/((x+1)(x-1)(x^2+2))`
Equating the coefficients from both the numerators we get,
A + B + C = 0........(1)
A - B + D = 1........(2)
2A + 2B - C = 0........(3)
2A - 2B - D= 0........(4)
Solving the above equations we get,
`A=1/6, B=-1/6, C=0, D=2/3`
Our Integral becomes
`intx^/((x+1)(x-1)(x^2+2))dx=1/(6(x-1))-1/(6(x+1))+2/(3(x^2+2))dx`
`=1/6log(x-1)-1/6log(x+1)+2/3xx1/sqrt2 tan^-1 (x/sqrt2)+C`
`=1/6[log(x-1)-log(x+1)+2sqrt2tan^-1 (x/sqrt2)]+C`
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