Question

How do you evaluate `\int \frac { 1} { x ^ { 3} + x ^ { 2} - 2x } d x`?

Answer 1

`int \ 1/(x^3+x^2-2x) \ dx = -1/2ln|x|+1/6ln|x+2|+1/3ln|x-1| + C`

Explanation:

We want to find:

`I = int \ 1/(x^3+x^2-2x) \ dx`

We can factorise the denominator to get:

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

And then we can expand the integrand into partial fractions:

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

`" " = (A(x+2)(x-1) +Bx(x-1)+Cx(x+2))/(x(x+2)(x-1))`

Leading to the identity:

`1 -= A(x+2)(x-1) +Bx(x-1)+Cx(x+2)`

To find the coefficients `A`, `B` and `C` we can use substitution:

Put `x= \ \ \ \ \ 0 => 1=A(2)(-1) \ \ \ \ => A=-1/2`
Put `x=-2 => 1=B(-2)(-3) => B=1/6`
Put `x= \ \ \ \ \ 1 => 1=C(1)(3) \ \ \ \ \ \ \ \ \ => C=1/3`

Hence the partial fraction decomposition of the integrand is:

`1/(x^3+x^2-2x) -= (-1/2)/x+(1/6)/(x+2)+(1/3)/(x-1)`

So we can write the integral as:

`I = int \ (-1/2)/x+(1/6)/(x+2)+(1/3)/(x-1) \ dx`

Which we can just integrate to get:

`I = -1/2ln|x|+1/6ln|x+2|+1/3ln|x-1| + C`