What is \(\rm \displaystyle\int \dfrac{dx}{x(x^n+1)}\) equal to?
Understanding the Integration Problem
We are asked to evaluate the indefinite integral of the function \(\dfrac{1}{x(x^n+1)}\) with respect to \(x\). This is a standard integral problem often encountered in calculus.
The integral we need to compute is:
\[ \int \dfrac{dx}{x(x^n+1)} \]
Step-by-Step Solution using Substitution and Partial Fractions
To solve this integral, we can use a combination of algebraic manipulation, substitution, and partial fraction decomposition.
Manipulating the Integrand
The presence of the \(x^n\) term suggests a substitution involving \(x^n\). If we let \(u = x^n\), then \(du = nx^{n-1} dx\). We can introduce an \(x^{n-1}\) term in the numerator by multiplying both the numerator and the denominator by \(x^{n-1}\):
\[ \dfrac{1}{x(x^n+1)} = \dfrac{x^{n-1}}{x \cdot x^{n-1} (x^n+1)} = \dfrac{x^{n-1}}{x^n (x^n+1)} \]
So, the integral becomes:
\[ \int \dfrac{x^{n-1}}{x^n (x^n+1)} dx \]
Applying Substitution
Let \(u = x^n\). Then, the differential \(du\) is given by:
\[ du = \dfrac{d}{dx}(x^n) dx = nx^{n-1} dx \]
From this, we have \(x^{n-1} dx = \dfrac{1}{n} du\).
Now, substitute \(u\) and \(du\) into the integral:
\[ \int \dfrac{1}{u(u+1)} \left(\dfrac{1}{n} du\right) = \dfrac{1}{n} \int \dfrac{du}{u(u+1)} \]
Partial Fraction Decomposition
The integrand \(\dfrac{1}{u(u+1)}\) can be decomposed into partial fractions. We want to find constants \(A\) and \(B\) such that:
\[ \dfrac{1}{u(u+1)} = \dfrac{A}{u} + \dfrac{B}{u+1} \]
Multiply both sides by \(u(u+1)\):
\[ 1 = A(u+1) + Bu \]
To find \(A\), set \(u=0\):
\[ 1 = A(0+1) + B(0) \implies 1 = A \implies A = 1 \]
To find \(B\), set \(u=-1\):
\[ 1 = A(-1+1) + B(-1) \implies 1 = 0 - B \implies B = -1 \]
So, the partial fraction decomposition is:
\[ \dfrac{1}{u(u+1)} = \dfrac{1}{u} - \dfrac{1}{u+1} \]
Integrating with Respect to \(u\)
Now, substitute the partial fractions back into the integral in terms of \(u\):
\[ \dfrac{1}{n} \int \left(\dfrac{1}{u} - \dfrac{1}{u+1}\right) du \]
Integrate term by term:
\[ \dfrac{1}{n} \left( \int \dfrac{1}{u} du - \int \dfrac{1}{u+1} du \right) \]
The integral of \(\dfrac{1}{u}\) is \(\ln|u|\), and the integral of \(\dfrac{1}{u+1}\) is \(\ln|u+1|\). So, we get:
\[ \dfrac{1}{n} (\ln|u| - \ln|u+1|) + C_1 \]
Using the logarithm property \(\ln a - \ln b = \ln \left(\dfrac{a}{b}\right)\), we have:
\[ \dfrac{1}{n} \ln\left|\dfrac{u}{u+1}\right| + C_1 \]
Substituting Back to \(x\)
Finally, substitute back \(u = x^n\):
\[ \dfrac{1}{n} \ln\left|\dfrac{x^n}{x^n+1}\right| + C_1 \]
Assuming \(x^n\) and \(x^n+1\) are positive (which is often the case in typical contexts for such problems), the absolute value signs can be dropped:
\[ \dfrac{1}{n} \ln\left(\dfrac{x^n}{x^n+1}\right) + c \]
Here, \(c\) represents the constant of integration.
Comparing with Options
Let's compare our result with the given options:
| Option | Expression |
|---|---|
| 1 | \( \rm \dfrac{1}{n} \ln \left(\dfrac{x^n}{x^n +1}\right)+c \) |
| 2 | \( \rm \ln \left(\dfrac{x^n+1}{x^n}\right)+c \) |
| 3 | \( \rm \ln \left(\dfrac{x^n}{x^n +1}\right)+c \) |
| 4 | \( \rm \dfrac{1}{n}\ln \left(\dfrac{x^n +1}{x^n}\right)+c \) |
Our calculated integral is \( \dfrac{1}{n} \ln\left(\dfrac{x^n}{x^n+1}\right) + c \), which exactly matches Option 1.
Revision Table: Key Steps for Integration \(\int \frac{dx}{x(x^n+1)}\)
| Step | Description | Action |
|---|---|---|
| 1 | Manipulate integrand | Multiply numerator and denominator by \(x^{n-1}\) to get \(\dfrac{x^{n-1}}{x^n(x^n+1)}\). |
| 2 | Apply substitution | Let \(u = x^n\), so \(du = nx^{n-1} dx\). Integral becomes \(\dfrac{1}{n} \int \dfrac{du}{u(u+1)}\). |
| 3 | Partial fractions | Decompose \(\dfrac{1}{u(u+1)}\) into \(\dfrac{1}{u} - \dfrac{1}{u+1}\). |
| 4 | Integrate in \(u\) | Integrate \(\dfrac{1}{n} \int \left(\dfrac{1}{u} - \dfrac{1}{u+1}\right) du\) to get \(\dfrac{1}{n}(\ln|u| - \ln|u+1|) + C\). |
| 5 | Simplify and substitute back | Use log properties: \(\dfrac{1}{n}\ln\left|\dfrac{u}{u+1}\right| + C\). Substitute \(u=x^n\) to get \(\dfrac{1}{n}\ln\left|\dfrac{x^n}{x^n+1}\right| + c\). |
Additional Information on Integration Techniques
This problem demonstrates the power of strategic manipulation and standard techniques like substitution and partial fraction decomposition in solving complex integrals.
- Substitution Method: This technique simplifies integrals by replacing the variable of integration with a new variable. The key is choosing a substitution \(u=g(x)\) such that the derivative \(du = g'(x) dx\) appears in the integrand, or can be easily introduced.
- Partial Fraction Decomposition: This method is used to integrate rational functions (ratios of polynomials). The rational function is broken down into simpler fractions that are easier to integrate, typically involving logarithms and inverse tangents. The denominator must be factorable into linear or irreducible quadratic factors.
- Logarithm Properties: Properties like \(\ln a - \ln b = \ln(a/b)\) and \(\ln a + \ln b = \ln(ab)\) are crucial for simplifying expressions involving logarithms that arise from integration.
Mastering these techniques is essential for solving a wide range of integration problems in calculus.
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