Problem 53 - A Complete Resource Book in Mathematics for JEE Main 2017

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).

Step 1: Simplify the Integral Expression

To solve the integral \( \int \frac{(x^2 - 1)}{(x^4 + 3x^2 + 1) \tan^{-1}\left( x + \frac{1}{x} \right)} \, dx \), let's start by simplifying the denominator \( (x^4 + 3x^2 + 1) \). Notice that this can be rewritten as: \((x^2 + 1)^2\). This simplifies our integral to: \( \int \frac{(x^2 - 1)}{(x^2 + 1)^2 \tan^{-1}(x + \frac{1}{x})} \, dx \).

Step 2: Choose a Substitution

Let's substitute \( u = \tan^{-1}\left(x + \frac{1}{x}\right) \). The derivative of \( u \) is \( du = \frac{1}{1 + (x + \frac{1}{x})^2} \left(1 - \frac{1}{x^2}\right) \, dx = \frac{x^2 - 1}{x^2 + 1} \, dx \). Therefore, \( dx = \frac{(x^2 + 1)du}{x^2 - 1} \).

Step 3: Substitute and Simplify the Integral

Substituting \( dx \) into the integral, we have: \( \int \frac{x^2 - 1}{(x^2 + 1)^2 \tan^{-1}(x + \frac{1}{x})} \cdot \frac{x^2 + 1}{x^2 - 1} \, du \). Simplifying, this becomes: \( \int \frac{1}{(x^2 + 1) \tan^{-1}(x + \frac{1}{x})} \, du \).

Step 4: Integrate

Now we can integrate with respect to \( u \): \( \int \frac{1}{u} \ du = \log |u| + c \). Substituting back for \( u \), we obtain: \( \log \left| \tan^{-1}(x + \frac{1}{x}) \right| + c \).

Step 5: Verify the Result

Compare the solution from Step 4 with the given options. The result corresponds to Option (A): \( \log \left|\tan^{-1}(x + \frac{1}{x})\right| + c \). Thus, the answer is option (A).

Key Concepts

Integration Techniques

Integration is a fundamental concept in calculus. It allows us to find the accumulated areas under curves or the total accumulation of a quantity. When facing a complex integral, choosing the right technique simplifies the task.

Several techniques are commonly used, such as:

• Direct Integration: Applying basic integration formulas directly to the integral.
• Substitution Method: Simplifying the integrand by substituting a part of it with a single variable.
• Integration by Parts: Used when the product of functions is involved, based on the product rule for differentiation.
• Partial Fraction Decomposition: Useful for rational functions, breaking them into simpler fractions.

In the given exercise, the integral appears complex due to the function inside, but a clever choice of substitution simplifies it greatly.

Substitution in Integrals

Substitution in integrals is a technique closely related to the chain rule for differentiation. It involves changing variables to simplify an integral, making it easier to solve.

For the provided integral, the substitution used is:

• Let: \( u = \tan^{-1}(x + \frac{1}{x}) \)
• Then the derivative \( du \) becomes: \( du = \frac{x^2-1}{x^2+1} \, dx \)
• Therefore, the substitution allows us to replace parts of the integral with \( u \) and \( du \).

This transformation simplifies the original integral into a more manageable form. The primary goal of substitution here was to simplify the integrand enough so that basic integration formulas could be applied.

Inverse Trigonometric Functions

Inverse trigonometric functions are the inverses of the trigonometric functions. They offer a way to find angles given trigonometric ratios. In calculus, they have a special significance due to their unique derivatives and integrals.

The function \( \tan^{-1}(x + \frac{1}{x}) \) was used as a substitution in the exercise. Its derivative is critical in simplifying the integration process and fits perfectly with the substitution method. Inverse trigonometric functions often appear in such problems since:

• Their derivatives are well-known and often simplify complex algebraic expressions.
• They arise naturally in problems involving angles and lengths.
• Their properties help convert intricate looking integrals to simpler forms.

Understanding these functions and their derivatives is important for mastering calculus integration problems.