Miscellaneous · Q2: $\int\frac{dx}{\sqrt{x+a}+\sqrt{x+b}}$

Basics you need

Solution

1. Domain, and why we rationalise

   Both roots are real when $x \ge -a$ and $x \ge -b$⁠. The denominator can only be 0 if both roots are 0, that is $x = -a$ and $x = -b$ at once, which is impossible when $a \ne b$⁠. So the integrand is defined for $x \ge \max(-a, -b)$⁠, and we work on the open interval $x > \max(-a, -b)$⁠, where both roots are positive.

   No standard formula has a sum of roots in the denominator, and a substitution such as $t = \sqrt{x+a}$ leaves the other root behind. Multiplying by the conjugate, however, turns the awkward denominator into a constant. This is the same idea as rationalising $\frac{1}{\sqrt3 + \sqrt2}$ in arithmetic.
   The square root $\sqrt p$ is real only for $p \ge 0$⁠, and $\sqrt p = 0$ only for $p = 0$⁠.

2. Multiply top and bottom by the conjugate

   The conjugate $\sqrt{x+a} - \sqrt{x+b}$ is not zero, because $x + a \ne x + b$ when $a \ne b$⁠. So we may multiply by it:

   $$\begin{aligned} f(x) &= \frac{\sqrt{x+a} - \sqrt{x+b}}{\big(\sqrt{x+a}+\sqrt{x+b}\big)\big(\sqrt{x+a}-\sqrt{x+b}\big)} \\ &= \frac{\sqrt{x+a} - \sqrt{x+b}}{(x+a) - (x+b)} \\ &= \frac{\sqrt{x+a} - \sqrt{x+b}}{a - b}. \end{aligned}$$
   The product in the denominator is $u^2 - v^2$ with $u = \sqrt{x+a}$⁠, $v = \sqrt{x+b}$⁠, and $u^2 = x + a$⁠, $v^2 = x + b$⁠. The $x$'s cancel, leaving the constant $a - b$⁠. This is exactly where $a \ne b$ is needed: otherwise we would be dividing by 0.
   Conjugates (Basic 1).

3. Integrate each square root with the power rule

   With $t = x + a$⁠, $dt = dx$⁠, and standard formula (i) with $n = \frac12$⁠:

   $$\int (x+a)^{1/2}\,dx = \frac{(x+a)^{3/2}}{3/2} = \frac23(x+a)^{3/2}.$$

   In the same way $\int (x+b)^{1/2}\,dx = \frac23(x+b)^{3/2}$⁠. Taking the constant $\frac{1}{a-b}$ outside:

   $$\begin{aligned} \int f(x)\,dx &= \frac{1}{a-b}\Big[\tfrac23(x+a)^{3/2} \\ &\qquad - \tfrac23(x+b)^{3/2}\Big] + C \\ &= \frac{2}{3(a-b)}\Big[(x+a)^{3/2} \\ &\qquad - (x+b)^{3/2}\Big] + C. \end{aligned}$$
   The power rule needs $n \ne -1$⁠; here $n = \frac12$⁠, so it applies and $n + 1 = \frac32$⁠. Dividing by $\frac32$ is multiplying by $\frac23$⁠. The answer is the same as the textbook's, term for term.
   Standard formula (i), Section 7.2; properties (III), (IV), Section 7.2.1; substitution, Section 7.3.1. Basics: the power rule

4. Check: differentiate the answer

   $\frac{d}{dx}(x+a)^{3/2} = \frac32(x+a)^{1/2}$⁠, and the same for $b$⁠. So

   $$\begin{aligned} &\frac{d}{dx}\Big(\tfrac{2}{3(a-b)}\big[(x+a)^{3/2} - (x+b)^{3/2}\big]\Big) \\ &= \frac{2}{3(a-b)}\cdot\frac32\big[\sqrt{x+a} - \sqrt{x+b}\big] \\ &= \frac{\sqrt{x+a} - \sqrt{x+b}}{a-b}. \end{aligned}$$

   Undo Step 2: $a - b = \big(\sqrt{x+a}+\sqrt{x+b}\big)\big(\sqrt{x+a}-\sqrt{x+b}\big)$⁠, so the last fraction is $\dfrac{1}{\sqrt{x+a}+\sqrt{x+b}}$⁠. ✓

   Cancelling the non-zero factor $\sqrt{x+a}-\sqrt{x+b}$ brings back the integrand exactly, so the antiderivative is right on the whole interval $x > \max(-a, -b)$⁠.
   Power rule and chain rule for derivatives, Chapter 5. Basics: integration undoes differentiation

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