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When trying to find the integral

$$I = \int (x\sqrt{2x + 5}) \mathrm{d}x$$

if we use the implicit substitution $u^2 = 2x + 5$, substituting into the integrand gives the sub-expression $\sqrt{u^2}$. We usually simplify it to $+u$ and proceed smoothly. My question is, why don't we consider $-u$ here?

EDIT:

I removed the explicit substitution, $u = 2x + 5$, let's focus on the implicit one.

reflectionalist
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2 Answers2

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By definition $\sqrt{x}$ indicates the positive (or principal) square root of $x$. If we want indicate all the two root we have to write explicitly $\pm\sqrt{x}$.

Enrico M.
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  • I understand this convention. But here we are dealing with not $\sqrt{x}$, but $\sqrt{x^2}$. As $x$ inside the square may be positive or negative, we should have $\sqrt{x^2} = \lvert x \rvert = \pm x$ (c.f. @expitt-i-1-z-0's comment). – reflectionalist Nov 23 '16 at 10:43
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  1. The implicit substitution $h(u) = 2x + 5,$ where $u$ is the new variable, is valid for each invertible function $h$.

  2. In this case, since $h(u)=u^2,$ it is tacit that $h$ has domain $[0,\infty).$

    This is why $\sqrt{u^2}\big(=|u|\big)$ can be simply expressed as $u.$

  3. Alternatively, we might specify $h$'s domain as $(-\infty,0],$ in which case $\sqrt{u^2}=-u$.

ryang
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