Question

How do you find `\int _ { 1} ^ { 2} \frac { x ^ { 2} - x - 16} { 32x - 2x ^ { 3} } d x`?

Answer 1

The integral has value `1/16ln(5/9) - 1/2ln2 ~~ -0.383`

Explanation:

We can factor `-2x^3 + 32x` as `-2x(x^2 - 16) = -2x(x + 4)(x - 4)`. We use partial fractions to integrate.

`A/(x + 4) + B/(x - 4) + C/(-2x) = (x^2 -x - 16)/(-2x(x + 4)(x -4))`

`A(-2x)(x - 4) + B(x + 4)(-2x) + C(x + 4)(x- 4) = x^2 - x - 16`

`A(-2x^2 + 8x) + B(-2x^2 - 8x) + C(x^2 - 16) = x^2 - x - 16`

`-2Ax^2 + 8Ax - 2Bx^2 - 8Bx + Cx^2 - 16C = x^2 - x - 16`

`(C - 2B - 2A)x^2 + (8A - 8B)x - 16C = x^2 - x - 16`

We can now say that

`{(C - 2B - 2A = 1), (8A - 8B = -1), (-16C = -16):}`

We can solve this as follows.

`-16C = -16 -> C = 1`

`1 - 2B - 2A = 1 -> -2B - 2A = 0 -> -2(B + A) = 0 -> A + B = 0`

This signifies that `A = -B`, so:

`8(-B) - 8B = -1`

`-8B - 8B = -1`

`-16B = -1`

`B = 1/16 -> A = -1/16`

We can now rewrite the definite integral.

`=int_1^2 1/(16(x - 4)) - 1/(16(x + 4)) - 1/(2x)dx`

We write as three separate integrals.

`=int_1^2 1/(16(x - 4))dx - int_1^2 1/(16(x +4))dx - int_1^2 1/(2x)dx`

`=[1/16ln|x - 4|]_1^2 - [1/16ln|x + 4|]_1^2 - [1/2ln|x|]_1^2`

`=1/16ln|-2| - 1/16ln|-3| - (1/16ln|6| - 1/16ln|5|) - (1/2ln|2| - 1/2ln|1|)`

`=1/16ln2 - 1/16ln3 - 1/16ln6 + 1/16ln5 - 1/2ln2 + 0`

`=1/16ln(5/9) - 1/2ln2`

Hopefully this helps!