Problem 96 - A Complete Resource Book in Mathematics for JEE Main

Question

\(\int \frac{d x}{x\left(x^{n}+1\right)}\) is equal to: (A) \(\frac{1}{n} \log \left(\frac{x^{\prime \prime}}{x^{n}+1}\right)+c\) (B) \(\frac{1}{n} \log \left(\frac{x^{n}+1}{x^{n}}\right)+c\) (C) \(\log \left(\frac{x^{n}}{x^{n}+1}\right)+c\) (D) none of these

Step-by-Step Solution

Answer

(B) \( \frac{1}{n} \log \left( \frac{x^n + 1}{x^n} \right) + c \) is the correct choice.

Step 1: Choose a Substitution

Let us start by using the substitution \( u = x^n \). Then we need to express \( dx \) and the other parts of the integral with respect to \( du \). Since \( u = x^n \), differentiating both sides gives \( du = n x^{n-1} dx \). Thus, \( dx = \frac{du}{n x^{n-1}} \).

Step 2: Express Integrand with Substitution

Our integral now becomes \( \int \frac{1}{x(u+1)} \cdot \frac{du}{n x^{n-1}} \). Simplifying, this integral becomes \( \frac{1}{n} \int \frac{1}{x^n (u+1)} du \). Since \( x^n = u \), our integrand simplifies to \( \frac{1}{n} \int \frac{1}{u(u+1)} du \).

Step 3: Partial Fraction Decomposition

We can decompose \( \frac{1}{u(u+1)} \) into partial fractions. Solving \( \frac{1}{u(u+1)} = \frac{A}{u} + \frac{B}{u+1} \) results in \( A(u+1) + Bu = 1 \). Setting \( u = 0 \) gives \( A = 1 \), and setting \( u = -1 \) gives \( B = -1 \). So, \( \frac{1}{u(u+1)} = \frac{1}{u} - \frac{1}{u+1} \).

Step 4: Integrate the Partial Fractions

Our integral now becomes \( \frac{1}{n} \left( \int \frac{1}{u} \, du - \int \frac{1}{u+1} \, du \right) \). Integrating each part separately, we get \( \frac{1}{n} ( \log |u| - \log |u+1| ) + C \). Using logarithmic properties, this simplifies to \( \frac{1}{n} \log \left( \frac{u}{u+1} \right) + C \).

Step 5: Substitute Back to Original Variable

Finally, recall that \( u = x^n \). Substitute back: \( \frac{1}{n} \log \left( \frac{x^n}{x^n + 1} \right) + C \).

Step 6: Match to the Given Options

Now let's match this answer \( \frac{1}{n} \log \left( \frac{x^n}{x^n + 1} \right) + C \) to the options provided. Option (B) matches this expression.

Key Concepts

Substitution Method

The substitution method is a powerful tool in integration. It simplifies integrals by introducing a new variable. By choosing a suitable substitution, we can transform a complicated integral into a simpler one.
In the original exercise, we use the substitution \( u = x^n \). This choice allows us to replace parts of the integrand and the differential \( dx \). When differentiating \( u = x^n \), we find that \( du = n x^{n-1} dx \). Solving for \( dx \) gives \( dx = \frac{du}{n x^{n-1}} \).
This process transforms the original integral into an integral in terms of \( u \), making it easier to handle. It's crucial in substitution to ensure that both the integrand and the differential are expressed in terms of \( u \). Choose substitutions that simplify the integral, particularly breaking down complex expressions into simpler, recognizable forms.

Partial Fraction Decomposition

Partial fraction decomposition is essential when dealing with rational expressions in integrals. It involves expressing a complicated fraction as a sum of simpler fractions.
In this exercise, once we have \( \int \frac{1}{u(u+1)} du \), we decompose it into partial fractions. We rewrite \( \frac{1}{u(u+1)} \) as \( \frac{A}{u} + \frac{B}{u+1} \).
By solving for \( A \) and \( B \) through substitution, we find \( A = 1 \) and \( B = -1 \). Thus, the expression becomes \( \frac{1}{u} - \frac{1}{u+1} \).
This method helps to break down the integral into simpler forms that we can integrate individually. It is particularly helpful for rational functions where the degree of the numerator is less than the degree of the denominator. Always remember, the goal of partial fractions is to facilitate easier integration.

Definite and Indefinite Integrals

Integration is a fundamental concept in calculus, with definite and indefinite integrals as its two types.
Indefinite integrals, like in this exercise, do not have specified limits of integration. Instead, they provide a family of functions, plus a constant \( C \). For example, \( \int f(x) \, dx = F(x) + C \).
Definite integrals, on the other hand, compute the net area under a curve over a specified interval \([a, b]\). Its result is a numerical value and not a function.
In the provided exercise, we find an indefinite integral. The process involves integrating and then adjusting the result with the constant \( C \). Remember, indefinite integrals provide general forms of functions, and the constant \( C \) accounts for all possible vertical translations of the antiderivative.