Exercise 7.1 · Q12: $\int \frac{x^3+3x+4}{\sqrt{x}}\,dx$

Basics you need

Solution

1. Where is the integrand defined?

   $\sqrt{x}$ is a real number only for $x \ge 0$⁠, and it sits in the denominator, so it must not be 0. Hence the integrand is defined exactly for

   $$x > 0.$$
   An antiderivative is always found on an interval where the integrand makes sense. Here that interval is $(0, \infty)$⁠, and the answer we get is valid there. It also tells us $x$ is positive, which is what lets us use the rules of exponents freely in the next step.
   The textbook's Note in Section 7.2: we usually do not mention the interval, "however, in any specific problem one has to keep it in mind".

2. Split the fraction into powers of $x$

   Divide $\sqrt{x} = x^{1/2}$ into each term of the numerator:

   $$\begin{aligned} \frac{x^3 + 3x + 4}{\sqrt{x}} &= \frac{x^3}{x^{1/2}} + \frac{3x}{x^{1/2}} + \frac{4}{x^{1/2}} \\ &= x^{3 - \frac12} + 3x^{1 - \frac12} + 4x^{-\frac12} \\ &= x^{5/2} + 3x^{1/2} + 4x^{-1/2}. \end{aligned}$$
   The power formula only applies to a single power $x^n$⁠. Splitting turns one hard quotient into three easy powers. The last term needs care: $\frac{4}{x^{1/2}}$ has $x$ in the denominator, so its exponent is negative, $-\frac12$⁠.
   $\frac{x^m}{x^n} = x^{m-n}$ and $\frac{1}{x^n} = x^{-n}$⁠. With $m = 3$⁠, $n = \frac12$⁠: $3 - \frac12 = \frac52$⁠; with $m = 1$⁠: $1 - \frac12 = \frac12$⁠. Basics: powers of x

3. Integrate term by term

   By Property V we may integrate each term separately, keeping the constants 3 and 4 in front. Then use the power formula with $n = \frac52$⁠, $\frac12$ and $-\frac12$⁠:

   $$\begin{aligned} \int x^{5/2}\,dx &= \frac{x^{7/2}}{7/2} = \frac{2}{7}x^{7/2}, \\ 3\int x^{1/2}\,dx &= 3\cdot\frac{x^{3/2}}{3/2} = 2x^{3/2}, \\ 4\int x^{-1/2}\,dx &= 4\cdot\frac{x^{1/2}}{1/2} = 8x^{1/2}. \end{aligned}$$
   Raise each power by 1: $\frac52 + 1 = \frac72$⁠, $\frac12 + 1 = \frac32$⁠, $-\frac12 + 1 = \frac12$⁠. Then divide by the new power. Dividing by a fraction means multiplying by its reciprocal, so $\div\frac72$ is $\times\frac27$⁠, $\div\frac32$ is $\times\frac23$⁠, and $\div\frac12$ is $\times 2$⁠. None of the exponents is $-1$⁠, so the formula applies to all three.
   Formula (i), Section 7.2: $\int x^n dx = \frac{x^{n+1}}{n+1} + C$⁠, $n \ne -1$⁠. Basics: the power rule

4. Assemble the answer with one constant

   $$\begin{aligned} &\int \frac{x^3+3x+4}{\sqrt{x}}\,dx \\ &\quad = \frac{2}{7}x^{7/2} + 2x^{3/2} + 8\sqrt{x} + C. \end{aligned}$$
   Each of the three integrals carries its own arbitrary constant, but the sum of three arbitrary constants is just one arbitrary constant, so we write a single $C$ (the textbook says so in the Note inside Example 2). We also wrote $x^{1/2}$ back as $\sqrt{x}$ to match the textbook's form.
   Properties III to V, Section 7.2.1. Lesson: integration as anti-differentiation

5. Check: differentiate the answer back

   Differentiate $F(x) = \frac{2}{7}x^{7/2} + 2x^{3/2} + 8x^{1/2} + C$ with the power rule $\frac{d}{dx}x^m = m x^{m-1}$⁠:

   $$\begin{aligned} F'(x) &= \tfrac27\cdot\tfrac72 x^{5/2} + 2\cdot\tfrac32 x^{1/2} \\ &\quad + 8\cdot\tfrac12 x^{-1/2} \\ &= x^{5/2} + 3x^{1/2} + 4x^{-1/2} \\ &= \frac{x^3 + 3x + 4}{\sqrt{x}}. \end{aligned}$$

   This is the integrand, for every $x > 0$⁠, so the answer is correct.

   By definition, $F$ is an integral of $f$ exactly when $F' = f$⁠. The last line reverses Step 2: multiply each term by $\frac{x^{1/2}}{x^{1/2}}$ and put everything over $\sqrt{x}$⁠. The constant $C$ differentiates to 0, which is why any value of $C$ works.
   Property I, Section 7.2.1: $\frac{d}{dx}\int f(x)\,dx = f(x)$⁠. Basics: integration undoes differentiation

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