Problem 20 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

\(\int \frac{[f(x) \cdot \phi(x)-\phi(x) \cdot \phi(x)]}{f(x) \cdot \phi(x)} \log \frac{f(x)}{\phi(x)} d x\) is equal to (A) \(\log \frac{\phi(x)}{f(x)}+k\) (B) \(\frac{1}{2}\left[\log \frac{\phi(x)}{f(x)}\right]^{2}+k\) (C) \(\frac{\phi(x)}{f(x)} \log \frac{\phi(x)}{f(x)}+k\) (D) none of these

Step-by-Step Solution

Answer

The correct answer is (B).

Step 1: Simplify the Integrand

The expression given in the integrand is \( \frac{[f(x) \cdot \phi(x) - \phi(x) \cdot \phi(x)]}{f(x) \cdot \phi(x)} \log \frac{f(x)}{\phi(x)} \).analyze the fraction part: \( \frac{f(x) \cdot \phi(x) - \phi(x) \cdot \phi(x)}{f(x) \cdot \phi(x)} = \frac{\phi(x)(f(x) - \phi(x))}{f(x) \cdot \phi(x)} = \frac{f(x) - \phi(x)}{f(x)} \). So, the integrand simplifies to \( \frac{f(x) - \phi(x)}{f(x)} \log \frac{f(x)}{\phi(x)} \).

Step 2: Recognize the Form of Integration

Recognize the form: the integral looks like \( \int u(x) \log(u(x)) \, dx \) where \( u(x) = \frac{f(x) - \phi(x)}{f(x)} \). This is the derivative of a certain form, indicating it might be an elementary integration problem where a substitution can fit.

Step 3: Consider Logarithmic Integration By Parts

Given that the simplified integrand is \( \log(v(x)) \cdot \frac{1}{v(x)} v'(x) \), suggest \( u = \log \frac{f(x)}{\phi(x)} \), and with integration by parts formulas, knowing that certain logarithmic expressions can integrate to specific forms reveals this could fit the formula \( \int \frac{1}{2} (\log(u(x)))^2 \).

Step 4: Compute the Integral

Using the recognition from Step 3, compute \( \int u(x) \log(u(x)) \, dx \) directly recognizes as \( \frac{1}{2} \left[ \log \frac{\phi(x)}{f(x)} \right]^2 \), completing the format presented in option B.

Step 5: Choose the Correct Answer

Given the derived integral matches one of the options: \( \frac{1}{2} \left[ \log \frac{\phi(x)}{f(x)} \right]^{2}+k \), it's clear that option B is the correct one.

Key Concepts

Integration by Parts

Integration by parts is a fundamental technique in calculus used to solve integrals involving the product of two functions. It essentially transforms a difficult integral into an easier one. The formula is derived from the product rule for differentiation and is given by:

• \(\int u \, dv = uv - \int v \, du\)

Here, \( u \) and \( dv \) are parts of the original integral, chosen in a way that makes \( \int v \, du \) simpler than the initial integral.

To apply this effectively, follow these steps:

• Select \( u \) such that its derivative \( du \) is simpler, and \( dv \) whose integral \( v \) is known.
• Differentiate \( u \) to get \( du \), and integrate \( dv \) to find \( v \).
• Substitute into the integration by parts formula to simplify the integral.
• Evaluate the resulting integrals.

Logarithmic Integration

Logarithmic integration specifically deals with integrals involving logarithmic functions. It often uses substitution and allows us to tackle complex integrals efficiently.

When dealing with integrals of the form \( \int u(x) \log u(x) \, dx \), where a logarithmic function depends on some variable expression, logarithmic integration can be particularly handy.

• First, identify a substitution that simplifies the expression under the logarithm.
• The goal is to express the integral in a form where standard techniques like integration by parts can apply.
• Using properties of logarithms, such integrals often resolve smoothly into more manageable forms.

In our exercise, we've applied this method to simplify the integrand before applying further techniques.

Integration Techniques

Various integration techniques exist to solve complex integrals, and choosing the right one depends on the integral's structure.

Some common techniques include:

• Substitution: Useful for integrals with functions and their derivatives. Substitution simplifies the integral into a basic form by changing the variable of integration.
• Partial Fractions: Applicable when dealing with the integration of rational functions. Breaking down a complex fraction into simpler ones can make integration straightforward.
• Integration by Parts: Effective for products of functions. It transforms the integral into a new form that is simpler or more recognizable.
• Trigonometric Substitution: Useful for integrals involving square roots of quadratic expressions. This converts the variable into a trigonometric function.

Mastering these techniques enables solving a wide range of integrals by recognizing their standard forms and applying the appropriate strategy. In the given exercise, logarithmic integration and parts worked together to achieve a solution.