Problem 53 - A Complete Resource Book in Mathematics for JEE Main

Question

\(\int \frac{\left(x^{2}-1\right)}{\left(x^{4}+3 x^{2}+1\right) \tan ^{-1}\left(x+\frac{1}{x}\right)} d x\) (A) \(\log \left|\tan ^{-1}\left(x+\frac{1}{x}\right)\right|+c\) (B) \(\log \left|\cot ^{-1}\left(x+\frac{1}{x}\right)\right|+c\) (C) \(2 \log \left|\tan ^{-1}\left(x+\frac{1}{x}\right)\right|+c\) (D) none of these

Step-by-Step Solution

Answer

The answer is (A) \(\log|\tan^{-1}(x+\frac{1}{x})| + c.\)

Step 1: Analyze the Integrand

The integrand is \( \frac{\left(x^{2}-1\right)}{\left(x^{4}+3x^{2}+1\right) \tan^{-1}\left(x+\frac{1}{x}\right)} \). Notice the \( x^4 + 3x^2 + 1 \) term in the denominator. This seems to factor or simplify with respect to \( x^2 \text{ and } \frac{1}{x^2} \).

Step 2: Factor or Replace

Set \( t = x + \frac{1}{x} \). Then, since \((x - \frac{1}{x})^2 = x^2 - 2 + \frac{1}{x^2}, \) it follows that an identity for \( x^2 + \frac{1}{x^2} = t^2 - 2. \) Therefore, the expression \( x^4 + 3x^2 + 1 \) can be replaced by \((x^2 + \frac{1}{x^2} + 2) \text{ which equals } t^2. \)

Step 3: Integrate Using Substitution

Now, consider the substitution strategy \( t = x + \frac{1}{x} \), then \( dt = \left(1 - \frac{1}{x^2} \right)dx, \) which simplifies to \( dt = \frac{x^2 - 1}{x^2} dx \). Thus, \( (x^2 - 1)dx = x^2 dt. \) The integral simplifies as \[ \int \frac{x^2 \, dt}{(t^2) \tan^{-1}(t)} = \int \frac{dt}{t \, \tan^{-1}(t)}. \]

Step 4: Integrate Resulting Expression

The integration \( \int \frac{1}{t \tan^{-1}(t)} \, dt \) can be found using the known integral formula \( \int \frac{1}{x \tan^{-1}(x)} \, dx = \log|\tan^{-1}(x)| + C. \)

Step 5: Final Answer

Thus, the integral evaluates to \( \log|\tan^{-1}(t)| + C \). Substituting \( t = x + \frac{1}{x} \), we have \( \log|\tan^{-1}(x + \frac{1}{x})| + C. \) Therefore, the answer matches option (A).

Key Concepts

Integration Techniques

Integration techniques are essential tools in calculus that allow us to solve complex integrals. These techniques often involve simplifying the integrands to make integration manageable. For instance, when facing an integral with a complicated fraction, one might simplify or decompose it before integration. These steps are crucial because they help us translate a challenging problem into a simpler one. Techniques like partial fractions, completing the square, or trigonometric identities frequently find use. Working through the integrand in pieces to arrive at a basic integration rule becomes the core strategy in handling intricate integrals.

In the given exercise, identifying that the denominator term could be associated with a substitution was key. Recognizing patterns, rearranging them into a known form, and later applying integration techniques results in efficient solutions.

Substitution Method

The substitution method in integration is much like the chain rule in differentiation. It involves changing variables to simplify the integral, generally replacing a composite part of the integrand with a single variable. This change often makes the integral easier or transforms it into a familiar form.

In the exercise, we employed substitution with the relation \( t = x + \frac{1}{x} \). This variable adjustment simplified the equation to integrate easily. Once substitution is made, we find the derivative to backtrack and adjust the integration accordingly. Here, by differentiating, we obtain \( dt = \left(1 - \frac{1}{x^2}\right)dx \), allowing us to replace and simplify the original integral effectively. Substitution is valuable because it transforms the integral's structure, making it resemble an easily solvable integral.

Inverse Trigonometric Functions

Inverse trigonometric functions, like \( \tan^{-1}(x) \), appear often when dealing with integration problems. These functions are significant due to their properties that link angles to values, which are instrumental when determining areas or solving equations within calculus.

The integral of inverse trigonometric functions follows specific patterns. For instance, the integral \( \int \frac{1}{x \tan^{-1}(x)} \, dx \) is known to result in \( \log|\tan^{-1}(x)| + C \). Recognizing when and how to apply these integral patterns saves considerable effort. In the exercise, this understanding allowed us to quickly integrate \( \int \frac{1}{t \tan^{-1}(t)} \, dt \). This directly led to the solution by using the known integral outcome of the inverse tangent function.

Mastering the integration of inverse trigonometric functions empowers students to tackle a wide range of calculable problems effortlessly. It also enriches understanding of how trigonometric concepts translate into calculus.