Miscellaneous · Q28: $\int_0^1\frac{dx}{\sqrt{1+x}-\sqrt{x}}$

Basics you need

Solution

1. The integrand is continuous on $[0, 1]$

   Both roots are defined for $x \ge 0$⁠. Since $1 + x > x \ge 0$ and the square root is increasing,

   $$\sqrt{1+x} > \sqrt{x} \quad\text{for every } x \ge 0 ,$$

   so the denominator $\sqrt{1+x} - \sqrt{x}$ is always positive and never zero. Hence the integrand is continuous on $[0, 1]$ (at $x = 0$ from the right, which is all a closed interval needs).

   Theorem 2 needs continuity on the closed interval. A difference in a denominator could vanish, so we must rule that out before doing anything else. Here it cannot: a larger number has a larger square root.
   If $0 \le q < p$ then $\sqrt{q} < \sqrt{p}$⁠. Lesson: definite integrals and the fundamental theorem

2. Rationalise: multiply by the conjugate

   Multiply the numerator and the denominator by $\sqrt{1+x} + \sqrt{x}$⁠, which is positive:

   $$\begin{aligned} \frac{1}{\sqrt{1+x} - \sqrt{x}} &= \frac{\sqrt{1+x} + \sqrt{x}}{(\sqrt{1+x})^2 - (\sqrt{x})^2} \\ &= \frac{\sqrt{1+x} + \sqrt{x}}{(1+x) - x} \\ &= \sqrt{1+x} + \sqrt{x} . \end{aligned}$$
   Multiplying top and bottom by the same non-zero number leaves a fraction unchanged. We pick the conjugate because the difference of squares makes both square roots disappear from the denominator. Here even the $x$ disappears, leaving 1. That is why the method fits this question so well: no fraction is left at all.
   $(p - q)(p + q) = p^2 - q^2$⁠, and $(\sqrt{a})^2 = a$ for $a \ge 0$⁠. Basics: splitting fractions

3. Find an antiderivative with the power rule

   $$\begin{aligned} &\int\big(\sqrt{1+x} + \sqrt{x}\big)\,dx \\ &= \int (1+x)^{1/2}\,dx + \int x^{1/2}\,dx \\ &= \tfrac23(1+x)^{3/2} + \tfrac23 x^{3/2} + C . \end{aligned}$$

   Call $F(x) = \frac23(1+x)^{3/2} + \frac23 x^{3/2}$⁠.

   The integral of a sum is the sum of the integrals (Section 7.2.1). Each term is a power with $n = \frac12$⁠: raise the power by 1 to $\frac32$ and divide by the new power $\frac32$⁠, that is, multiply by $\frac23$⁠. For $(1+x)^{1/2}$ the substitution $u = 1 + x$ has $du = dx$⁠, so the shift changes nothing in the rule.
   Standard formula (i), Section 7.2, with $n = \frac12$⁠. Basics: powers of x and the power rule

4. Check: differentiate $F$ back

   $$\begin{aligned} F'(x) &= \tfrac23\cdot\tfrac32(1+x)^{1/2} + \tfrac23\cdot\tfrac32 x^{1/2} \\ &= \sqrt{1+x} + \sqrt{x} . \end{aligned}$$

   This is the rationalised integrand, which equals the original one by Step 2 ✓. It holds for every $x \ge 0$⁠.

   Differentiating the answer back confirms both the power rule and the factor $\frac23$⁠. The chain rule factor from $1 + x$ is $\frac{d}{dx}(1+x) = 1$⁠.
   Basics: integration undoes differentiation

5. Apply Theorem 2 and simplify

   $F(1)$ $= \frac23\cdot 2^{3/2} + \frac23\cdot 1$ $= \frac23\cdot 2\sqrt2 + \frac23$ $= \frac{4\sqrt2 + 2}{3}$⁠.

   $F(0) = \frac23\cdot 1^{3/2} + \frac23\cdot 0 = \frac23$⁠.

   $$\begin{aligned} I &= F(1) - F(0) \\ &= \frac{4\sqrt2 + 2}{3} - \frac23 = \frac{4\sqrt2}{3} . \end{aligned}$$

   As a decimal, $I \approx 1.885618$⁠.

   $2^{3/2} = 2\cdot 2^{1/2} = 2\sqrt2$⁠. The two $\frac23$ terms cancel: one comes from $(1+x)^{3/2}$ at $x = 0$⁠, the other from $x^{3/2}$ at $x = 1$⁠. Size check: by Step 2 the integrand rises from $1 + 0 = 1$ at $x = 0$ to $\sqrt2 + 1 \approx 2.414$ at $x = 1$⁠, so the integral over an interval of length 1 must lie between 1 and 2.414. It does.
   Theorem 2, Section 7.8.3. Lesson: definite integrals

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