Question

What is `int_(2)^(8) (x-1)/(x^3+x^2)dx`?

Answer 1

`int_2^8 (x - 1)/(x^3 + x^2) = 2ln|4/3| - 3/8`

Explanation:

The denominator can be factored as

`x^2(x + 1)`

Now using partial fractions, we have:

`(Ax + B)/x^2 + C/(x+ 1) = (x - 1)/(x(x + 1))`

`Ax^2 + Bx + Ax + B + Cx^2 = x - 1`

We now write a system of equation `{(A + C = 0), (A + B = 1), (B = -1):}`

If we solve, we get:

`A = 2, B = -1, C = -2`

Hence the partial fraction decomposition is

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

The integral becomes:

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

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

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

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

`I = 2ln|x| + x^-1 - 2ln|x + 1| + C`

Which can be written as

`I = 2ln|(x)/(x + 1)| + x^-1 + C`

We now evaluate the definite integral.

`I = 2ln|8/9| + 1/8 - (2ln|2/3| + 1/2)`

`I = 2ln|8/9| - 2ln|2/3| + 1/8 - 1/2`

`I = 2(ln|(8/9)/(2/3)|) - 3/8`

`I = 2ln|4/3| - 3/8`

Hopefully this helps!