Problem 24 - A Complete Resource Book in Mathematics for JEE Main

Question

\(\int \frac{\left(x^{2}-2\right) d x}{\left(x^{4}+5 x^{2}+4\right) \tan ^{-1}\left(\frac{x^{2}+2}{x}\right)}\) is (A) \(\log \left|\tan ^{-1} \sqrt{x+2}\right|+C\) (B) \(\log \left|\tan ^{-1}\left(x+\frac{2}{x}\right)\right|+C\). (C) \(\sin ^{-1}\left(\frac{x+2}{x}\right)+C\) (D) \(\tan ^{-1}\left(\frac{x+2}{x}\right)+C\)

Step-by-Step Solution

Answer

The solution is option (B): \(\log \left|\tan^{-1}\left(x + \frac{2}{x}\right)\right| + C\).

Step 1: Identify the Integral Components

We need to evaluate \(int \frac{(x^2 - 2) \, dx}{(x^4 + 5x^2 + 4) \cdot \tan^{-1}\left(\frac{x^2 + 2}{x}\right)}\). We notice that the denominator has the component \(\tan^{-1}\left(\frac{x^2 + 2}{x}\right)\), suggesting a potential derivative simplification.

Step 2: Simplify the Expression

Focus on the expression \(\tan^{-1}\left(\frac{x^2 + 2}{x}\right)\). Let \(u = \tan^{-1}\left(\frac{x^2 + 2}{x}\right)\). Then, \(\frac{du}{dx} = \frac{d}{dx}\left(\tan^{-1}\left(\frac{x^2 + 2}{x}\right)\right)\).

Step 3: Compute the Derivative

Using the chain rule and the derivative of \(\tan^{-1}(v)\), where \(v = \frac{x^2 + 2}{x}\), it follows that \(\frac{du}{dx} = \frac{1}{1 + \left(\frac{x^2 + 2}{x}\right)^2} \cdot \frac{d}{dx}\left(\frac{x^2 + 2}{x}\right)\).

Step 4: Calculate \(\frac{d}{dx}\left(\frac{x^2 + 2}{x}\right)\)

Differentiate \(\frac{x^2 + 2}{x}\) using the quotient rule: \( \frac{d}{dx}\left(\frac{x^2 + 2}{x}\right) = \frac{x \cdot 2x - (x^2 + 2) \cdot 1}{x^2} = \frac{2x^2 - x^2 - 2}{x^2} = \frac{x^2 - 2}{x^2}\).

Step 5: Integrate Using Substitution

Using the derivative from step 4, notice that the numerator of the integral is \(x^2 - 2\). The derivative \(\frac{d}{dx}\left(\tan^{-1}\left(\frac{x^2 + 2}{x}\right)\right)\) is equivalent to the form \(\frac{(x^2 - 2)}{(x^4 + 5x^2 + 4)} \cdot \tan^{-1}\left(\frac{x^2 + 2}{x}\right)\). This simplifies the integration to \(\int \frac{du}{u}\).

Step 6: Finalize the Solution

Solving \(\int \frac{du}{u}\) yields \(\log |u| + C\). Substituting back \(u = \tan^{-1}\left(\frac{x^2 + 2}{x}\right)\), we get the solution \(\log \left|\tan^{-1}\left(\frac{x^2 + 2}{x}\right)\right| + C\).

Key Concepts

Trigonometric Substitution

In calculus, certain integrals can be complicated to solve directly. This is where trigonometric substitution comes into play. Its goal is to use trigonometric identities to simplify the integration process, especially when dealing with expressions having square roots of quadratic polynomials.

One common scenario for using trigonometric substitution is when integrals involve expressions like \( \sqrt{a^2 - x^2} \), \( \sqrt{x^2 - a^2} \), or \( \sqrt{x^2 + a^2} \). For example, to simplify an integral involving \( \sqrt{a^2 - x^2} \), the substitution \( x = a \sin \theta \) is used. This substitution exploits the identity \( 1 - \sin^2 \theta = \cos^2 \theta \) to eliminate the square root in the expression.

After substitution, the integrand is expressed entirely in terms of the trigonometric functions and the new variable, often resulting in easier to solve integrals. Once the integral is evaluated, the result is converted back to the original variable using inverse trigonometric functions. This step secures the integration result in terms of the initial variable.

Inverse Trigonometric Functions

Inverse trigonometric functions are essential in calculus, especially when dealing with antiderivatives that involve trigonometric substitutions. These functions include \( \sin^{-1}(x) \), \( \cos^{-1}(x) \), and \( \tan^{-1}(x) \).

The notation \( \tan^{-1}(x) \), for instance, represents the angle whose tangent is \( x \). This function is commonly used in calculus problems involving angles or trigonometric substitutions, as seen in our exercise where \( u = \tan^{-1} \left( \frac{x^2 + 2}{x} \right) \).

Inverse trigonometric functions also have derivatives that are quite useful. For example, the derivative of \( \tan^{-1}(x) \) is \( \frac{1}{1 + x^2} \), an identity applied in differentiating expressions that involve these functions. When an integration result involves an inverse trigonometric function, it often helps in back-substituting values to solve the original integral.

Integration by Substitution

Also known as "u-substitution," integration by substitution is a pivotal technique in calculus that simplifies integration by transforming the variable of integration. It's similar to the Chain Rule for differentiation but applied in reverse for integrals.

Consider the integral \( \int f(g(x)) g'(x) \, dx \). If we let \( u = g(x) \), then \( du = g'(x) \, dx \). Rewriting the integral in terms of \( u \) gives \( \int f(u) \, du \), often a simpler integral to evaluate.

In our exercise, the substitution method is crucial. By recognizing that the derivative of the inner function in the denominator matches the integral's numerator \( x^2 - 2 \), it facilitates a straightforward transformation to \( \int \frac{du}{u} \), which is easier to solve. After integrating to \( \log |u| + C \), back-substituting \( u = \tan^{-1} \left( \frac{x^2 + 2}{x} \right) \) completes the solution.