Question

1. Write the form of the partial fraction decomposition of the rational expression. Do not solve for the constants.

8x − 4/x(x2 + 1)2
2. Use partial fractions to find the indefinite integral. (Remember to use absolute values where appropriate. Use C for the constant of integration).
2x3 - 16x2 - 39x + 20/x2 - 8x - 20 x dx
3. Use partial fractions to find the indefinite integral. (Remember to use absolute values where appropriate. Use C for the constant of integration).
x2/x4 - 42x2 - 343 x dx

Answer 1

Explanation:

1.

To write the form of the partial fraction decomposition of the rational expression:

We have:

`\mathbf{(8x-4)/(x(x^2+1)^2)= (A)/(x)+(Bx+C)/(x^2+1)+(Dx+E)/((x^2+1)^2)}`

2.

Using partial fraction decomposition to find the definite integral of:

`(2x^3-16x^2-39x+20)/(x^2-8x-20)dx`

By using the long division method; we have:

`x^2-8x-20 | (2x)/(2x^3-16x^2-39x+20 )`

`- 2x^3 -16x^2-40x`

`x+ 20`

So;

`(2x^3-16x^2-39x+20)/(x^2-8x-20)= 2x+(x+20)/(x^2-8x-20)`

By using partial fraction decomposition:

`(x+20)/((x-10)(x+2))= (A)/(x-10)+(B)/(x+2)`

`= (A(x+2)+B(x-10))/((x-10)(x+2))`

x + 20 = A(x + 2) + B(x - 10)

x + 20 = (A + B)x + (2A - 10B)

Now; we have to relate like terms on both sides; we have:

A + B = 1 ; 2A - 10 B = 20

By solvong the expressions above; we have:

`A = (5)/(2)`
`B = (3)/(2)`

Now;

`(x+20)/((x-10)(x+2)) = (5)/(2(x-10)) + (3)/(2(x+2))`

Thus;

`(2x^3-16x^2-39x+20)/(x^2-8x-20)= 2x + (5)/(2(x-10))+ (3)/(2(x+2))`

Now; the integral is:

`\int (2x^3-16x^2-39x+20)/(x^2-8x-20) \ dx = \int \begin {bmatrix} 2x + (5)/(2(x-10))+ (3)/(2(x+2)) \end {bmatrix} \ dx`

`\mathbf(3)/(2) In`

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