Q.Integrate the function $f'(ax+b)\,[f(ax+b)]^n$

The key idea is the Chain Rule — the integrand is exactly the derivative of a composite function.

1. Let $u = f(ax+b)$. Then $\frac{du}{dx} = f'(ax+b) \cdot a$, so $f'(ax+b)\,dx = \frac{du}{a}$.
2. The integral becomes $\int u^n \cdot \frac{du}{a} = \frac{1}{a} \int u^n \, du$. …

When you see an integrand like $f'(ax+b)\,[f(ax+b)]^n$, your first instinct should be: this is screaming for the chain rule in reverse. The chain rule tells us that if we differentiate a composite function $F(g(x))$, we get $F'(g(x)) \cdot g'(x)$. Here, the "outer" function is something like $u^{n+1}$ (since the power $n$ suggests we want to increase the exponent by 1), and the "inner" function is $f(ax+b)$. The factor $f'(ax+b)$ is exactly the derivative of the inner function, except for the constant $a$ that comes from differentiating $ax+b$.

Let’s unpack that carefully.

1. Identify the inner function and its derivative.

   Let $u = f(ax+b)$. Then $\frac{du}{dx} = f'(ax+b) \cdot a$ (by the chain rule: derivative of $f$ times derivative of $ax+b$, which is $a$). So $du = a \cdot f'(ax+b)\,dx$, or equivalently $f'(ax+b)\,dx = \frac{du}{a}$.

2. Rewrite the integral in terms of $u$.

   The integrand $f'(ax+b)\,[f(ax+b)]^n\,dx$ becomes $u^n \cdot \frac{du}{a}$. That is:

3. Integrate with respect to $u$.

   The power rule for integration gives $\int u^n \, du = \frac{u^{n+1}}{n+1} + C$, provided $n \neq -1$. (If $n = -1$, the integral becomes $\int \frac{1}{u}\,du = \log|u| + C$, a separate case.)

4. Substitute back.

   Replace $u$ with $f(ax+b)$:

   $\frac{1}{a} \cdot \frac{[f(ax+b)]^{n+1}}{n+1} + C.$ …

Method: Reverse chain rule for $\int f'(ax+b)\,[f(ax+b)]^{n}\,dx$

Use this whenever a composite function is raised to a power and multiplied by (a piece of) its own derivative — recognise it as the derivative of a higher power.

Steps

Step 1: Substitute the inner function.

Let $u=f(ax+b)$. By the chain rule $du=a\,f'(ax+b)\,dx$, so $f'(ax+b)\,dx=\dfrac{du}{a}$. The extra constant $a$ comes from differentiating $ax+b$ — this is the factor most often forgotten.

Step 2: Apply the power rule in $u$. …

Common Mistakes

Mistake 1: Forgetting the factor $a$ from $ax+b$.

Why it's wrong: $\tfrac{d}{dx}f(ax+b)=a\,f'(ax+b)$, so $f'(ax+b)\,dx=\tfrac{du}{a}$; missing the $a$ leaves the answer off by a factor of $a$. Correct approach: include $\tfrac1a$ in the antiderivative.

Mistake 2: Ignoring the $n=-1$ case. …