Problem 60 - A Complete Resource Book in Mathematics for JEE Main

Question

\(\int \frac{(x-1) d x}{(x+1) \sqrt{x^{3}+x^{2}+x}}=k \tan ^{-1} \sqrt{\frac{x^{2}+x+1}{x}}+c^{\prime}\) where \(k=\) (A) \(]\) (B) 2 (C) 4 (D) none of these

Step-by-Step Solution

Answer

The value of \(k\) is 2.

Step 1: Simplify the Integral

We start by simplifying the expression under the integral sign. Observe that the denominator can be rewritten as a product of two terms: \((x+1)\sqrt{x}\sqrt{x^2+x+1}\). This separation will help later in integration.

Step 2: Make a Substitution

Use the substitution \(u = \sqrt{\frac{x^2 + x + 1}{x}}\), which implies \(u^2 = \frac{x^2 + x + 1}{x}\) or \(u^2 x = x^2 + x + 1\). This simplifies to \(x(u^2 - 1) = x^2 + x\), suggesting it might be possible to integrate with respect to \(u\) after finding \(\frac{du}{dx}\).

Step 3: Differentiate and Solve for dx

Differentiate \(u^2\) with respect to \(x\) to get \[2u\frac{du}{dx} = \frac{-1 \cdot (x^2+x+1) - (2x+1) \cdot x}{x^2}\] which simplifies to \(2u\frac{du}{dx} = \frac{-2x - 1}{x^2}\). This will assist in changing the variable from \(x\) to \(u\).

Step 4: Substituting Back

Re-writing the integral in terms of \(u\): \[\int \frac{(x-1)dx}{(x+1) \sqrt{x^{3}+x^{2}+x}} = \int \frac{-u^2 dx}{2u(\frac{-2x-1}{x^2})}\] Simplify to encounter \[\frac{1}{2}\int \frac{x}{x+1} du\].

Step 5: Apply Inverse Trigonometric Function Integration

Notice that the function resembles an inverse tangent after substituting, suggesting \[\int u \, du \rightarrow \tan^{-1} u + C\] Thus, \[\int \frac{(x-1) dx}{(x+1)\sqrt{x^{3}+x^{2}+x}} = k \tan^{-1}\sqrt{\frac{x^{2}+x+1}{x}} + c\].

Step 6: Determine the Constant k

Comparing with the given result, the solution is written as \[k \tan^{-1}\sqrt{\frac{x^{2}+x+1}{x}}\]. Knowing the simplified integral resulted in a factor change, determine \(k = 2\) to satisfy the comparison transformation.

Key Concepts

Trigonometric Substitution

When dealing with complex integrals, trigonometric substitution often simplifies the process. This technique involves replacing a part of the integrand with a trigonometric function. In this example, the substitution specifically helps when dealing with the square root expressions, typical in integrals resembling derivatives of inverse trigonometric functions.
Trigonometric substitution is useful when a square root of a sum or difference of squared terms is present. For instance, terms like \(\sqrt{a^2 - x^2}\) suggest using \(x = a \sin(\theta)\), which simplifies the square root to \(a \cos(\theta)\). This transforms a potentially complicated algebraic expression into a manageable trigonometric integral.
In our problem, the substitution helps express complicated roots in a form more conducive to integration with respect to trigonometric identities. Understandably, this substitution works hand-in-hand with the substitution method to complete the integration process successfully.

Inverse Trigonometric Functions

Inverse trigonometric functions play a significant role in integration, especially when the integral's structure resembles derivatives of these functions. In our example, the result leads us to consider the arctangent or \(\tan^{-1}\) function.
Recognizing integral forms that simplify to inverse trigonometric functions can streamline solving complex integrals. For instance:

• The integral \(\int \frac{1}{1+x^2} dx\) simplifies to \(\tan^{-1}(x) + C\).
• Similarly, \(\int \frac{1}{\sqrt{1-x^2}} dx\) results in \(\sin^{-1}(x) + C\).

Returning to the exercise, after making appropriate substitutions, the resulting structure was likened to inverse tangent integration, thus linking the problem to the form \(\tan^{-1}\). Moreover, identifying how inverse trigonometric functions emerge during integration is imperative to completing problems efficiently.

Substitution Method

The substitution method acts as a bridge to express integrals in a simpler form. In managing integrals, substitution involves replacing a complex expression with a single variable. This method simplifies calculations significantly, reducing errors and helping isolate integral parts that are easier to evaluate.
In this exercise, initially identifying \(u = \sqrt{\frac{x^2 + x + 1}{x}}\) as a substitution targets the unwieldy denominator. This substitution aimed to untangle a complex root and polynomial interaction that might otherwise make direct integration difficult or less obvious.
By substituting and differentiating to find \(\frac{du}{dx}\), our method turns a potentially perplexing expression into a format that matches simpler known integrals, often involving inverse trigonometric derivatives. Recognizing when and how to substitute significantly enhances solving complex integrals while preventing unnecessary algebraic manipulations.