Example 2: $\int \frac{x^3 - 1}{x^2}\,dx$⁠, $\int (x^{2/3} + 1)\,dx$⁠, $\int (x^{3/2} + 2e^x - \frac1x)\,dx$

Basics you need

Part (i): $\int \frac{x^3 - 1}{x^2}\,dx$

1. Split the fraction into powers of $x$

   For $x \ne 0$⁠, divide each term of the numerator by $x^2$⁠:

   $$\frac{x^3 - 1}{x^2} = \frac{x^3}{x^2} - \frac{1}{x^2} = x - x^{-2}.$$
   There is no formula for a quotient, but there is one for each power of $x$⁠. A fraction with a single term in the denominator can always be split term by term, because $\frac{a - b}{c} = \frac ac - \frac bc$⁠. The condition $x \ne 0$ is needed for the integrand to exist at all.
   Index laws. Basics: splitting fractions

2. Integrate term by term

   By Property V, $\int (x - x^{-2})\,dx = \int x\,dx - \int x^{-2}\,dx$⁠. Use formula (i) with $n = 1$ and $n = -2$ (neither is $-1$⁠):

   $$\begin{aligned} \int x\,dx &= \frac{x^{1+1}}{1+1} + C_1 = \frac{x^2}{2} + C_1, \\ \int x^{-2}\,dx &= \frac{x^{-2+1}}{-2+1} + C_2 = -\frac1x + C_2. \end{aligned}$$
   For $n = -2$⁠: the new power is $-2 + 1 = -1$⁠, and we divide by $-1$⁠, giving $\frac{x^{-1}}{-1} = -\frac1x$⁠. Each integral brings its own constant; the book writes them as $C_1$ and $C_2$ to make the point that they are independent.
   Standard formula (i), Section 7.2; Property V, Section 7.2.1. Basics: the power rule

3. Subtract and merge the constants

   $$\begin{aligned} \int \frac{x^3 - 1}{x^2}\,dx &= \left(\frac{x^2}{2} + C_1\right) - \left(-\frac1x + C_2\right) \\ &= \frac{x^2}{2} + \frac1x + (C_1 - C_2). \end{aligned}$$

   $C_1 - C_2$ is just another arbitrary constant; call it $C$⁠. So

   $$\int \frac{x^3 - 1}{x^2}\,dx = \frac{x^2}{2} + \frac1x + C, \qquad x \ne 0.$$
   Subtracting $-\frac1x$ gives $+\frac1x$⁠: watch this sign. Since $C_1$ and $C_2$ can each be any real number, so can $C_1 - C_2$⁠, and nothing is lost by writing a single $C$⁠. From here on the book writes only one constant in the final answer.
   Note after Example 2(i) in the book. Lesson: integration as anti-differentiation

4. Check: differentiate the answer

   $$\begin{aligned} \frac{d}{dx}\left(\frac{x^2}{2} + \frac1x\right) &= x - \frac{1}{x^2} \\ &= \frac{x^3 - 1}{x^2}. \ \checkmark \end{aligned}$$
   $\frac{d}{dx}x^{-1} = -x^{-2}$⁠. Putting the two terms over the common denominator $x^2$ returns the original integrand, valid for every $x \ne 0$⁠.

Part (ii): $\int (x^{2/3} + 1)\,dx$

1. Split and integrate each term

   $$\int \left(x^{\frac23} + 1\right)dx = \int x^{\frac23}\,dx + \int dx.$$

   Formula (i) with $n = \frac23$⁠: the new power is $\frac23 + 1 = \frac53$⁠, so

   $$\int x^{\frac23}\,dx = \frac{x^{\frac53}}{\frac53} = \frac35\,x^{\frac53},$$

   and $\int dx = x$ (the special case $n = 0$⁠).

   Dividing by the fraction $\frac53$ is the same as multiplying by its reciprocal $\frac35$⁠. The "1" in the integrand is $x^0$⁠, and $\int x^0\,dx = \frac{x^1}{1} = x$⁠.
   Standard formula (i) and $\int dx = x + C$⁠, Section 7.2. Basics: the power rule

2. Write the answer with one constant

   $$\int \left(x^{\frac23} + 1\right)dx = \frac35\,x^{\frac53} + x + C.$$

   Reading $x^{2/3}$ as $\left(\sqrt[3]{x}\right)^2$⁠, the integrand is defined for every real $x$⁠, and the answer holds for every real $x$⁠.

   Cube roots exist for negative numbers, so $x^{2/3}$ and $x^{5/3}$ make sense for all $x$⁠. The check in the next step works at negative $x$ as well (it is tested at $x = -8$ below).

3. Check: differentiate the answer

   $$\frac{d}{dx}\left(\frac35\,x^{\frac53} + x\right) = \frac35\cdot\frac53\,x^{\frac23} + 1 = x^{\frac23} + 1. \ \checkmark$$

   For example at $x = -8$⁠: $\sqrt[3]{-8} = -2$⁠, so the integrand is $(-2)^2 + 1 = 5$⁠, and the derivative formula gives the same $5$⁠.

   The power rule $\frac{d}{dx}x^{5/3} = \frac53 x^{2/3}$ brings down the factor $\frac53$⁠, which cancels the $\frac35$ exactly. This cancellation is why we divided by the new power.

Part (iii): $\int (x^{3/2} + 2e^x - \frac1x)\,dx$

1. Where the integrand exists

   $x^{3/2} = \left(\sqrt{x}\right)^3$ needs $x \ge 0$⁠, and $\frac1x$ needs $x \ne 0$⁠. So we work on $x > 0$⁠.

   An answer is only meaningful where the integrand is defined. Here the square root rules out negative $x$⁠, and the $\frac1x$ rules out 0.
   Domains of powers and of $\frac1x$⁠. Basics: powers of x

2. Split into three standard integrals

   $$\begin{aligned} &\int \left(x^{\frac32} + 2e^x - \frac1x\right)dx \\ &= \int x^{\frac32}dx + 2\int e^x\,dx - \int \frac{dx}{x}. \end{aligned}$$
   Property V lets us split the sum and take the constant 2 outside its integral. Each of the three pieces is now on the list of standard formulae.
   Property V, Section 7.2.1. Lesson: integration as anti-differentiation

3. Apply formulae (i), (xi) and (xii)

   $$\begin{aligned} \int x^{\frac32}dx &= \frac{x^{\frac32 + 1}}{\frac32 + 1} = \frac25\,x^{\frac52}, \\ 2\int e^x\,dx &= 2e^x, \\ \int \frac{dx}{x} &= \log|x|. \end{aligned}$$

   Therefore

   $$\begin{aligned} &\int \left(x^{\frac32} + 2e^x - \frac1x\right)dx \\ &= \frac25\,x^{\frac52} + 2e^x - \log|x| + C. \end{aligned}$$
   $\frac32 + 1 = \frac52$⁠, and dividing by $\frac52$ means multiplying by $\frac25$⁠. $e^x$ is its own derivative, so it is its own anti derivative. The $\frac1x$ term is the exception $n = -1$ of the power formula and gives a logarithm. Since $x > 0$ here, $\log|x| = \log x$⁠; the book keeps the modulus to match formula (xii).
   Standard formulae (i), (xi), (xii), Section 7.2. Basics: logarithms and 1/x

4. Check: differentiate the answer

   $$\begin{aligned} &\frac{d}{dx}\left(\frac25\,x^{\frac52} + 2e^x - \log|x|\right) \\ &= \frac25\cdot\frac52\,x^{\frac32} + 2e^x - \frac1x \\ &= x^{\frac32} + 2e^x - \frac1x. \ \checkmark \end{aligned}$$
   Each term differentiates back to the term it came from, which is exactly what term-by-term integration promised.

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