Question

Prove the formula $$ \int \frac{g(x) f^{\prime}(x)-f(x) g^{\prime}(x)}{g^{2}(x)} d x=\frac{f(x)}{g(x)}+C $$

Short answer

The formula is valid as the integrand is the derivative of \( \frac{f(x)}{g(x)} \).

Step-by-step solution

Identify the Formula

We need to prove that \( \int \frac{g(x) f^{\prime}(x)-f(x) g^{\prime}(x)}{g^{2}(x)}\, d x = \frac{f(x)}{g(x)} + C \). This can be approached by using the Quotient Rule for derivatives and the Fundamental Theorem of Calculus.

Use Quotient Rule

Recall the derivative of a quotient \( \frac{f(x)}{g(x)} \) is given by the Quotient Rule: \( \frac{d}{dx} \left( \frac{f(x)}{g(x)} \right) = \frac{g(x)f'(x) - f(x)g'(x)}{g^2(x)} \). This closely resembles our integrand.

Recognize the Integrand as a Derivative

Notice that the integrand \( \frac{g(x)f'(x) - f(x)g'(x)}{g^2(x)} \) is exactly the derivative of \( \frac{f(x)}{g(x)} \). Thus, integrating the derivative \( \frac{d}{dx} \left( \frac{f(x)}{g(x)} \right) \) should yield \( \frac{f(x)}{g(x)} \) plus an integration constant.

Apply the Fundamental Theorem of Calculus

Apply the Fundamental Theorem of Calculus, which states that the integral of a derivative function returns the original function plus a constant of integration. So, \( \int \frac{g(x)f'(x) - f(x)g'(x)}{g^2(x)} \, dx = \frac{f(x)}{g(x)} + C \). Thus, this proves the given formula.

Key concepts

Quotient Rule

The quotient rule is a powerful technique used in calculus. It helps us find the derivative of a ratio where one function is divided by another function. To use the quotient rule, remember the following formula:
\[ \frac{d}{dx}\left(\frac{f(x)}{g(x)}\right) = \frac{g(x)f'(x) - f(x)g'(x)}{g^2(x)} \]

This formula is crucial because it allows us to differentiate complex functions involving division. When applying the quotient rule, ensure that:

• \(f(x)\) is the numerator function.
• \(g(x)\) is the denominator function.
• Both \(f'(x)\) and \(g'(x)\) are their respective derivatives.

In our exercise, the quotient rule helps identify the integrand expression as the derivative of \(\frac{f(x)}{g(x)}\). Understanding this connection is the key to connecting differentiation with the subsequent integration step, simplifying the given integral.

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus is central to the relationship between differentiation and integration. It comprises two parts and links the concept of the derivative of a function with its integral.

The theorem states:

• If \(F(x)\) is an antiderivative of a function \(f(x)\) over an interval, then the integral of \(f(x)\) from \(a\) to \(b\) is \(F(b) - F(a)\).
• If we have \(F'(x) = f(x)\), then integrating \(f(x)\) yields \(F(x)\) plus a constant \(C\).

Applying this theorem, if the integrand is a derivative of a known function, integrating it returns that function itself up to a constant. In our solution, recognizing that the integrand was \( \frac{d}{dx}\left(\frac{f(x)}{g(x)}\right) \), allows us to straightforwardly conclude that integrating yields \( \frac{f(x)}{g(x)} + C \), which confirms the validity of the exercise's formula.

Integration of Derivatives

Integration of derivatives may at first seem redundant. However, it's a powerful calculus concept. It proves that by following certain steps, one can reverse the effect of differentiation, effectively undoing it.
Consider the relationship:
If \( \frac{d}{dx}(F(x)) = f(x) \), and when \( f(x) \) is integrated, it reproduces \( F(x) \) with an integration constant \( C \).

In our given problem,

• We have an integrand \( \frac{g(x)f'(x) - f(x)g'(x)}{g^2(x)} \).
• This is the derivative of \( \frac{f(x)}{g(x)} \)

Once the integrand was recognized as a derivative, integrating it returned the original function divided, \( \frac{f(x)}{g(x)} \), plus the constant \( C \).
This seamless transition between differentiation and integration allows calculus to flexibly tackle a variety of function types, ultimately making calculus one of the most powerful mathematical tools in studying change.