Question

How do you evaluate the integral `int 1/(x^2+2x+1)dx` from -1 to 0?

Answer 1

There is a problem as `x!=-1`

Explanation:

Let's factorise the denominator
`x^2+2x+1=(x+1)^2`
We use `intx^ndx=x^(n+1)/(n+1) (x!=-1)`
There is a problem as `x!=-1`

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