`int (x^2+12x+12)/(x^3-4x)dx `

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

`(x^2+12x+12)/(x^3-4x) =(x^2+12x+12)/(x(x-2)(x+2))`

Then, express it as sum of fractions.

`(x^2+12x+12)/(x(x-2)(x+2)) = A/x + B/(x-2) + C/(x+2)`

To determine the values of A, B and C, multiply both sides by the LCD of...

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`int (x^2+12x+12)/(x^3-4x)dx `

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

`(x^2+12x+12)/(x^3-4x) =(x^2+12x+12)/(x(x-2)(x+2))`

Then, express it as sum of fractions.

`(x^2+12x+12)/(x(x-2)(x+2)) = A/x + B/(x-2) + C/(x+2)`

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

`x(x-2)(x+2)*(x^2+12x+12)/(x(x-2)(x+2)) = (A/x + B/(x-2) + C/(x+2))*x(x-2)(x+2)`

`x^2+12x+12=A(x-2)(x+2) +Bx(x+2)+Cx(x-2)`

Then, assign values to x in which either x, x-2 or x+2 will become zero.

So, plug-in x=0 to get the value of A.

`0^2+12(0)+12=A(0-2)(0+2)+B(0)(0+2)+C(0)(0-2)`

`0+0+12=A(-4)+B(0)+C(0)`

`12=-4A`

`-3=A`

Also, plug-in x=2 to get the value of B.

`2^2+12(2)+12=A(2-2)(2+2)+B(2)(2+2)+C(2)(2-2)`

`4+24+12=A(0)+B(8)+C(0)`

`40=8B`

`5=B`

And subsitute x=-2 to get the value of C.

`(-2)^2 + 12(-2)+12=A(-2-2)(-2+2)+B(-2)(-2+2)+C(-2)(-2-2)`

`4-24+12=A(0)+B(0)+C(8)`

`-8=8C`

`-1=C`

So the partial fraction decomposition of the integral is

`int (x^2+12x+12)/(x^3-4x)dx`

`= int (x^2+12x+12)/(x(x-2)(x+2))dx`

`= int(-3/x +5/(x-2)-1/(x+2))dx`

Then, express it as three integrals.

`= int-3/xdx + int 5/(x-2)dx - int 1/(x+2)dx`

`= -3int 1/xdx + 5int 1/(x-2)dx - int 1/(x+2)dx`

To take the integral, apply the formula `int 1/u du =ln|u|+C` .

`=-3ln|x| + 5ln|x-2|-ln|x+2|+C`

**Therefore,`int (x^2+12x+12)/(x^3-4x)dx=-3ln|x| + 5ln|x-2|-ln|x+2|+C`.**