`int x^2/(x^4-2x^2-8) dx` Use partial fractions to find the indefinite integral

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`int x^2/(x^4-2x^2-8)dx`

To solve using partial fraction method, the denominator of the integrand should be factored.

`x^2/(x^4-2x^2-8)=x^2/((x-2)(x+2)(x^2+2))`

If the factor in the denominator is linear, its partial fraction has a form `A/(ax+b)` .

If the factor is quadratic, its partial fraction is in the form `(Ax+B)/(ax^2+bx+c)` .

So, expressing the integrand as sum of fractions, it becomes:

`x^2/((x-2)(x+2)(x^2+2)) = A/(x-2) + B/(x + 2) + (Cx + D)/(x^2+2)`

To determine the values of A, B, C and D, multiply both sides by the LCD of the fractions present

`(x-2)(x+2)(x^2+2)*x^2/((x-2)(x+2)(x^2+2)) = (A/(x-2) + B/(x + 2) + (Cx + D)/(x^2+2))*(x-2)(x+2)(x^2+2)`

`x^2=A(x+2)(x^2+2) + B(x-2)(x^2+2) + (Cx+D)(x-2)(x+2)`

Then, assign values to x...

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