Problem 96
Question
\(\int \frac{d x}{x\left(x^{n}+1\right)}\) is equal to : (A) \(\frac{1}{n} \log \left(\frac{x^{n}}{x^{x}+1}\right)+c\) (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
Verified Answer
The integral is equal to (A): \( \frac{1}{n} \log \left(\frac{x^{n}}{x^{n}+1}\right)+c \).
1Step 1: Identify the Integral
We are given the integral \( \int \frac{d x}{x\left(x^{n}+1\right)} \). Our goal is to find an antiderivative of this expression.
2Step 2: Use Substitution
Let's perform a substitution to simplify the integral.Let \( u = x^n \). Then, \( du = n x^{n-1} dx \). This implies \( dx = \frac{du}{n x^{n-1}} \).Plug this back into the integral, we have:\[ \int \frac{1}{x(x^n + 1)} \cdot \frac{du}{n x^{n-1}} = \frac{1}{n} \int \frac{1}{x^n(x^n + 1)} du \]We know \( x^n = u \), so the integral becomes:\[ \frac{1}{n} \int \frac{1}{u(u + 1)} du \]
3Step 3: Partial Fraction Decomposition
We decompose \( \frac{1}{u(u+1)} \) into partial fractions.Assume \( \frac{1}{u(u+1)} = \frac{A}{u} + \frac{B}{u+1} \).Solving this equation gives the system:\[ 1 = A(u+1) + Bu \]\[ 1 = Au + A + Bu \]From which, comparing coefficients, we find:\[ A + B = 0 \quad \text{and} \quad A = 1 \]Thus: \( A = 1 \) and \( B = -1 \).So, \( \frac{1}{u(u+1)} = \frac{1}{u} - \frac{1}{u + 1} \).
4Step 4: Integrate Using Partial Fractions
Substitute these partial fractions back into the integral:\[ \frac{1}{n} \int \left( \frac{1}{u} - \frac{1}{u+1} \right) du \]Integrate term by term:\[ \frac{1}{n} \left( \ln |u| - \ln |u+1| \right) + C \]This simplifies to:\[ \frac{1}{n} \ln \left| \frac{u}{u+1} \right| + C \]
5Step 5: Substitute Back the Original Variable
Recall that \( u = x^n \), substitute \( u \) back into the expression:\[ \frac{1}{n} \ln \left| \frac{x^n}{x^n + 1} \right| + C \]This matches option (A), considering absolute values and adjustment of constants.
Key Concepts
Partial Fraction DecompositionSubstitution MethodAntiderivative
Partial Fraction Decomposition
Partial fraction decomposition is a technique used to break down complex rational expressions into simpler fractions that are easier to integrate or differentiate. In this exercise, we encountered the expression \[ \frac{1}{u(u+1)} \].
This expression might seem complicated, but by using partial fraction decomposition, we can express it in terms of simpler fractions. Let's assume the original expression can be rewritten as:
- \( \frac{A}{u} + \frac{B}{u+1} \), with constants \( A \) and \( B \).
To find these constants, consider \( 1 = A(u+1) + Bu \). Expanding and simplifying this will help determine \( A \) and \( B \).
In our exercise, comparing coefficients resulted in \( A = 1 \) and \( B = -1 \). Thus, we can write:- \( \frac{1}{u(u+1)} = \frac{1}{u} - \frac{1}{u+1} \).
Using partial fraction decomposition allows us to reduce complex integrals to simpler ones, making integration feasible.
This expression might seem complicated, but by using partial fraction decomposition, we can express it in terms of simpler fractions. Let's assume the original expression can be rewritten as:
- \( \frac{A}{u} + \frac{B}{u+1} \), with constants \( A \) and \( B \).
To find these constants, consider \( 1 = A(u+1) + Bu \). Expanding and simplifying this will help determine \( A \) and \( B \).
In our exercise, comparing coefficients resulted in \( A = 1 \) and \( B = -1 \). Thus, we can write:- \( \frac{1}{u(u+1)} = \frac{1}{u} - \frac{1}{u+1} \).
Using partial fraction decomposition allows us to reduce complex integrals to simpler ones, making integration feasible.
Substitution Method
The substitution method is a powerful tool in integration, especially when dealing with functions that include composition. It simplifies integrals by changing variables, turning a challenging integral into one that is straightforward.
In the given exercise, \( u = x^n \) was chosen as the substitution variable.
This helps simplify the original integral \( \int \frac{d x}{x(x^{n}+1)} \). Here's why:
- When \( u = x^n \), the differential becomes \( du = nx^{n-1}dx \), solving this for \( dx \) gives us \( dx = \frac{du}{n x^{n-1}} \).
- Replacing \( dx \) in the integral using this expression adjusts the integrand, easing further simplification.
By substituting back appropriately, the expression transforms, making the integral more manageable and ready for partial fraction decomposition.
In the given exercise, \( u = x^n \) was chosen as the substitution variable.
This helps simplify the original integral \( \int \frac{d x}{x(x^{n}+1)} \). Here's why:
- When \( u = x^n \), the differential becomes \( du = nx^{n-1}dx \), solving this for \( dx \) gives us \( dx = \frac{du}{n x^{n-1}} \).
- Replacing \( dx \) in the integral using this expression adjusts the integrand, easing further simplification.
By substituting back appropriately, the expression transforms, making the integral more manageable and ready for partial fraction decomposition.
Antiderivative
Finding an antiderivative is key when dealing with integrals. An antiderivative of a function is a function whose derivative gives the original function. In integration, we're essentially searching for this antiderivative.
For our problem, after applying substitution and partial fractions, the integral simplifies to:- \( \frac{1}{n} \left( \ln |u| - \ln |u+1| \right) + C \),
where \( C \) is the constant of integration.
This result is an antiderivative of the rewritten function in terms of \( u \). The last step involves back-substituting the original variable \( x \), giving:- \( \frac{1}{n} \ln \left| \frac{x^n}{x^n + 1} \right| + C \).
It's essential to check the correctness of the antiderivative by differentiating it, ensuring it corresponds to the original integrand. In integration, finding the antiderivative requires careful application of these techniques.
For our problem, after applying substitution and partial fractions, the integral simplifies to:- \( \frac{1}{n} \left( \ln |u| - \ln |u+1| \right) + C \),
where \( C \) is the constant of integration.
This result is an antiderivative of the rewritten function in terms of \( u \). The last step involves back-substituting the original variable \( x \), giving:- \( \frac{1}{n} \ln \left| \frac{x^n}{x^n + 1} \right| + C \).
It's essential to check the correctness of the antiderivative by differentiating it, ensuring it corresponds to the original integrand. In integration, finding the antiderivative requires careful application of these techniques.
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