A simple integral with one question 
Question is: For $x$ equals $4$ and $9$, why is $t$ not $\pm2$ and $\pm3$ but just $2$ and $3$ ?
 A: You can use the negative square root if you want;  it all works out the same way in the end.  If we denote the positive square root of $x$ by $\sqrt{x}$, we can substitute $t = -\sqrt{x}$ instead.  We still have $t^2 = x$, so $2 t \, dt = dx $
as before.  So the integral becomes
\begin{align*}
\int_{-2}^{-3} \frac{2 t \, dt}{-t -1} &= 2 \int_{-3}^{-2} \frac{t}{t + 1} dt \\
&= 2 \left[ \int_{-3}^{-2} dt  - \int_{-3}^{-2} \frac{dt}{t+1} \right] \\
&= 2 \Big[ t - \ln |t+1| \Big]_{-3}^{-2} \\
&= 2 \Big[ -2 - (-3) - \left( \ln(1) - \ln(2) \right)\Big] \\
&= 2 + 2 \ln 2.
\end{align*}
The steps look slightly different, but it works out to be exactly the same result.  This is an important lesson in general:  there is frequently more than one possible substitution that allows you to solve an integral.  
A: By convention, $\sqrt{a}$ represents the non-negative square root, or the principal square root, of $x$. Hence, the only case in which there are two opposite solutions is when you have $\pm\sqrt{a}$. (Note the extra $\pm$ sign.) Hence is important to note that $x = \sqrt{a}$ (one non-negative solution). should not be confused with $x^2 = a \iff \vert x\vert = \sqrt{a} \iff x = \pm\sqrt{a}$ (two solutions). So, for instance, $\sqrt{4} = \color{blue}{+}2$ and $\sqrt{9} = \color{blue}{+}3$.
A: Because under standard and generally accepted notation,
$$ \forall x\ge0;\sqrt x =|x^{\frac12}|$$
That is to say, the square root is used in this case to mean the positive square root.
