`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 in which either `x-2`,`x+2` or`x^2+2`will become zero.

So plug-in `x=2` to get the value of A.

`2^2=A(2+2)(2^2+2) + B(2-2)(2^2+2) + (C*2+D)(2-2)(2+2)`

`4=A(24)+B(0)+(2C+D)(0)`

`4=24A`

`1/6=A`

Plug-in`x=-2` to get the value of B.

`(-2)^2=A(-2+2)((-2)^2+2) + B(-2-2)((-2)^2+2) + (C(-2)+D)(-2-2)(-2+2)`

`4=A(0)+B(-24) + (-2C+D)(0)`

`4=-24B`

`-1/6=B`

To solve for D, plug-in the values of A and B. Also, set the value of x to zero.

`0^2 =1/6(0+2)(0^2+2) + (-1/6)(0-2)(0^2+2) + (C(0)+D)(0-2)(0+2)`

`0=2/3+2/3-4D`

`0=4/3-4D`

`4D=4/3`

`D=1/3`

To solve for C, plug-in the values of A, B and D. Also, assign any value to x. Let it be  `x=1`.

`1^2=1/6(1+2)(1^2+2)+( -1/6)(1-2)(1^2+2) + (C(1)+1/3)(1-2)(1+2)`

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

`1=3/2+1/2-3C -1`

`1=1-3C`

`0=-3C`

`0=C`

So the partial fraction decomposition of the integrand is:

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

`=intx^2/((x-2)(x+2)(x^2+2))dx`

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

Expressing it as three integrals, it becomes

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

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

For the first and second integral, apply the formula `int 1/u du = ln|u|+C` .

And for the third integral, the formula is`int 1/(u^2+a^2)du =1/a tan^(-1)(u/a)+C` .

`=1/6ln|x-2| - 1/6ln|x+2| + 1/3*1/sqrt2 tan^(-1)(x/sqrt2)+C`

`=1/6ln|x-2| - 1/6ln|x+2| + 1/(3sqrt2) tan^(-1)(x/sqrt2)+C`

`=1/6ln|x-2| - 1/6ln|x+2| + sqrt2/6 tan^(-1)((xsqrt2)/2)+C`

Therefore, `int x^2/(x^4-2x^2-8)dx=1/6ln|x-2| - 1/6ln|x+2| + sqrt2/6 tan^(-1)((xsqrt2)/2)+C` .

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