Miscellaneous · Q17: $\int f'(ax+b)\,[f(ax+b)]^n\,dx$

Basics you need

Solution

1. Choose the substitution $t = f(ax+b)$

   $$t = f(ax+b) ,\qquad \frac{dt}{dx} = a\,f'(ax+b) ,$$ $$f'(ax+b)\,dx = \frac{dt}{a} .$$
   The integrand is "something to a power" times "the derivative of that something" (apart from the constant $a$⁠). This is the pattern substitution is made for. Dividing by $a$ is allowed because $a \ne 0$⁠.
   Chain rule (Chapter 5); Section 7.3.1. Basics: substitution

2. Integrate in $t$ (for $n \ne -1$⁠)

   $$\begin{aligned}\int f'(ax+b)\left[f(ax+b)\right]^n dx &= \int t^n\cdot\frac{dt}{a}\\ &= \frac1a\cdot\frac{t^{n+1}}{n+1} + C .\end{aligned}$$
   The constant $\frac1a$ comes outside (Property (IV), Section 7.2.1), and the power formula needs $n + 1 \ne 0$⁠.
   Standard formula (i), Section 7.2. Basics: powers of x

3. Return to $x$

   $$\begin{aligned}&\int f'(ax+b)\left[f(ax+b)\right]^n dx\\ &= \frac{\left[f(ax+b)\right]^{n+1}}{a(n+1)} + C .\end{aligned}$$
   Replace $t$ by $f(ax+b)$⁠.

4. Check: differentiate the answer

   Chain rule twice: outer $u^{n+1}$⁠, middle $u = f(v)$⁠, inner $v = ax+b$⁠:

   $$\begin{aligned}&\frac{d}{dx}\,\frac{\left[f(ax+b)\right]^{n+1}}{a(n+1)}\\ &= \frac{(n+1)\left[f(ax+b)\right]^{n}\cdot f'(ax+b)\cdot a}{a(n+1)}\\ &= f'(ax+b)\left[f(ax+b)\right]^n\ \checkmark\end{aligned}$$
   The factors $a$ and $n+1$ produced by differentiating cancel the denominator exactly, which is why the answer carries $a(n+1)$ underneath.
   Basics: integration undoes differentiation

5. The excluded case $n = -1$

   Then the integrand is $\frac{f'(ax+b)}{f(ax+b)}$ and the same substitution gives

   $$\int\frac{f'(ax+b)}{f(ax+b)}\,dx = \frac1a\int\frac{dt}{t} = \frac1a\log\left|f(ax+b)\right| + C ,$$

   valid wherever $f(ax+b) \ne 0$⁠.

   The formula of Step 3 would divide by $n + 1 = 0$⁠, so it cannot be used; $\int\frac{dt}{t}$ is a logarithm instead (standard formula (xii)). The question does not mention this case, but a complete answer states it.
   Basics: logarithms and 1/x

6. Conditions, and the textbook's answer

   • $a \ne 0$ (we divided by $a$⁠). If $a = 0$⁠, the integrand is the constant $f'(b)\left[f(b)\right]^n$ and the integral is that constant times $x$⁠, plus $C$⁠.
   • $n \ne -1$ for the formula of Step 3 (Step 5 covers $n = -1$⁠).
   • The power must make sense: for a positive integer $n$⁠, any $x$⁠; for a negative integer $n$⁠, only where $f(ax+b) \ne 0$⁠; for a fractional $n$⁠, usually only where $f(ax+b) > 0$⁠.

   The key gives $\frac{\left[f(ax+b)\right]^{n+1}}{a(n+1)} + C$⁠, exactly our answer (the key leaves these conditions unstated).

   An answer is valid on any interval where both the integrand and the answer are defined and the derivative check of Step 4 holds; these bullet points describe exactly those intervals.

7. A numerical example

   Take $f = \sin$⁠, $a = 2$⁠, $b = 1$⁠, $n = 3$⁠. The question becomes $\int\cos(2x+1)\sin^3(2x+1)\,dx$⁠, and the formula gives

   $$\frac{\sin^4(2x+1)}{2\cdot 4} + C = \frac{\sin^4(2x+1)}{8} + C .$$

   From 0 to 1 this predicts $\frac{\sin^4 3 - \sin^4 1}{8} \approx -0.0626214205$⁠, and numerical integration of $\cos(2x+1)\sin^3(2x+1)$ from 0 to 1 also gives $-0.0626214205$⁠.

   A general formula is easiest to trust after one concrete case. The negative value is fine: the integrand is negative on most of $[0, 1]$⁠, because $2x+1$ runs from 1 to 3 radians and $\cos$ is negative beyond $\frac{\pi}{2} \approx 1.57$⁠.

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