double integral getting different results I am trying to calculate the double integral $$\lim_{b \to 0^+} \int_{b}^1 \int_b^1 \frac{y-x}{(y+x)^3}dydx$$ If you plug this into wolfram, you get $-\frac{1}{2}$ and if you plug it into symbolab you get $\frac{1}{2}$ I will show you my steps, I just want to make sure I got the right answer.
$$\lim_{b \to 0^+} \int_{b}^1 \int_b^1 \frac{y-x}{(y+x)^3}dydx=\lim_{b \to 0^+} \int_{b}^1 \int_b^1 \frac{y+x}{(y+x)^3}-\frac{2x}{(y+x)^3}dydx$$
$$=\lim_{b \to 0^+} \int_{b}^1 \frac{-1}{(1+x)^2}dx=\lim_{b \to 0^+} \frac{1}{1+x}\Big|_b^1=\frac{-1}{2}$$ I just wanted to verify because these two different websites are giving me different answers.

I may be wrong but I think the correct result is $0$.
Where I think you got it wrong is in the second equality, where you exchange limit and the integral for one term.
Here I sketched some more detailed computations:
\begin{align}
  \lim_{b \to 0}
    \int_b^1 \int_b^1 \frac{y-x}{(y+x)^3}dydx
{}={} &
  \lim_{b \to 0}
     \int_b^1 \int_b^1
        \left(\frac{y+x}{(y+x)^3}-\frac{2x}{(y+x)^3}\right)dydx
\\
{}={} &
  \lim_{b \to 0}
     \int_b^1 
        \left(
          -\frac{1}{y+x}\Bigg|_{y=b}^{y=1}
          +
          \frac{2x}{2(y+x)^2}\Bigg|_{y=b}^{y=1}
         \right)dx
\\
{}={} &
  \lim_{b \to 0}
     \int_b^1 
        \left(
          \frac{1}{b+x}
          -\frac{1}{1+x}
          +
          \frac{x}{(1+x)^2}
          -\frac{x}{(b+x)^2}
         \right)dx
\\
{}={} &
  \lim_{b \to 0}
     \int_b^1 
        \left(
          \frac{-1}{(1+x)^2}
          +\frac{b}{(b+x)^2}
         \right)dx
\\
{}={} &
  \lim_{b\to0}
  \left(
    \frac{1}{1+x}
    -\frac{b}{b+x}
    \Bigg|_{x=b}^{x=1}
  \right)
\\
{}={} &
  \lim_{b\to0}
  \left(
    \frac{1}{2}
    -\frac{1}{1+b}
    +\frac{b}{2b}
    -\frac{b}{b+1}
  \right)
\\
{}={} &
   0
\end{align}
