Question

Express the integrands as sum of partial fractions and evaluate the integral. The integral from -1 to 0 of (x^3 dx)/ (x^2-2x+1)?

I get how to do partial fractions, it's just that the limits are throwing me off.

Answer 1

Using polynomial long division to simplify the integrand, evaluating the integrals we know, and then decomposing the final integral and adding up all the evaluations, we get: `2-ln(8)`

Explanation:

The degree of the numerator is greater than the degree of the denominator; therefore, we must first divide the denominator into the numerator, otherwise we cannot decompose into partial fractions .

Using polynomial long division, we get:

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

Rewrite the integral:
`int_-1^0[x+2+(3x-2)/(x^2-2x+1)]dx`

Break it apart:
`int_-1^0xdx + int_-1^0 2dx + int_-1^0(3x-2)/(x^2-2x+1)dx`

We can integrate the first two terms easily and evaluate them from `-1` to `0`: :

`(x^2/2+2x)` evaluated from `-1` to `0` becomes

`0-(1/2-2)=3/2`

We're left with

`int_-1^0(3x-2)/(x^2-2x+1)dx`

We can factor the denominator into `(x-1)^2`:

`int_-1^0(3x-2)/(x-1)^2dx`

Decompose `(3x-2)/(x-1)^2`:

`(3x-2)/(x-1)^2=A/(x-1)+B/(x-1)^2`

By adding on the other side and setting numerators equal, we get:

`(3x-2)=A(x-1)+B`

Find `B` by setting `x` equal to `1`:

`B=1`

There is no value of `x` we can plug in to get `A`, so let's multiply out the right side and set coefficients equal:

`3x-2=Ax-A+B`

`A=3`

Rewrite our integral with our partial fractions:

`int_-1^0(3x-2)/(x-1)^2dx=int_-1^0 3/(x-1)dx + int_-1^0 1/(x-1)^2dx`

These are elementary integrals, taking them, we find we must evaluate `3ln|x-1|` from `-1` to `0`, giving us:

`3ln|-1|-3ln|-2|=3ln(1)-3ln(2)=0-3ln(2)=-3ln(2)=-ln(2^3)=-ln(8)`

And we must evaluate `-1/(x-1)` from `-1` to `0`, giving us:

`1-(1/2)=1/2`

We can now add up all of our evaluated definite integrals:

`1/2+3/2-ln(8)=2-ln(8)`