Since the equation provided by the problem is a transcedental equation, you may either solve it using graphical method, or you need to prove that the value of the function `f(x) = ln(1 + x^2) + x - 1` at `x = -1` is positive/negative and the value of the function `f(x) =` `ln(1 + x^2) + x - 1` at `x = 1` is negative/positve, hence, the function has the zero value at `x = c in (-1,1)` .

You need to evaluate `f(x)` at `x = -1` , hence, you need to replace -1 for x in equation of function, such that:

`f(-1) = ln(1 + (-1)^2) - 1 - 1 => f(-1) = ln 2 - 2`

Since `ln 2 ~~ 0.692` yields:

`f(-1) = 0.692 - 2 = -1.308 < 0`

You need to evaluate `f(x)` at `x = 1` , hence, you need to replace `1` for `x` in equation of function, such that:

`f(1) = ln(1 + 1^2) + 1 - 1 => f(1) = ln 2 ~~ 0.692 > 0`

**Hence, evaluating the values of the function at `x =-1` and `x = 1` yields that from between the negative value at `x = -1` and the positive value at `x = 1` , the function has a zero value at `x = c in (-1,1)` , thus the equation `ln(1+x^2) + x - 1 = 0` has one solution in `(-1,1)` .**