What is \(\smallint \frac{{dx}}{{x{{\left( {1 + In\;x} \right)}^n}}}\) equal to (n ≠ 1) ?
Understanding the Integral Problem
The problem asks us to evaluate the indefinite integral: \(\smallint \frac{{dx}}{{x{{\left( {1 + In\;x} \right)}^n}}}\) where \(n \ne 1\). This integral looks like a good candidate for the substitution method, also known as u-substitution.
Applying the Substitution Method for Integration
The structure of the integrand \(\frac{1}{{x{{\left( {1 + In\;x} \right)}^n}}}\) suggests that the term \(1 + \ln x\) might be a suitable choice for substitution because its derivative involves \(\frac{1}{x}\), which is also present in the integrand.
Let's use the substitution method:
- Choose a substitution: Let \(u = 1 + \ln x\).
- Find the differential \(du\): Differentiate both sides with respect to \(x\): \(\frac{{du}}{{dx}} = \frac{d}{{dx}}(1 + \ln x) = 0 + \frac{1}{x} = \frac{1}{x}\) Rearranging, we get \(du = \frac{1}{x} dx\). This matches the \( \frac{{dx}}{x} \) part of our original integral.
- Rewrite the integral in terms of \(u\): Substitute \(u = 1 + \ln x\) and \(du = \frac{1}{x} dx\) into the original integral: \(\smallint \frac{{dx}}{{x{{\left( {1 + In\;x} \right)}^n}}} = \smallint \frac{1}{{{{\left( {1 + In\;x} \right)}^n}}} \cdot \frac{{dx}}{x} = \smallint \frac{1}{{{u^n}}} du\) We can rewrite \( \frac{1}{{{u^n}}} \) as \(u^{-n}\). So the integral becomes: \(\smallint {u^{ - n}} du\)
- Integrate with respect to \(u\): Now we integrate \(u^{-n}\) using the power rule for integration, which states \(\smallint y^k dy = \frac{{y^{k+1}}}{{k+1}} + C\), provided \(k \ne -1\). In our case, \(y = u\) and \(k = -n\). The problem states \(n \ne 1\), which means \(-n \ne -1\), so the power rule is applicable. \(\smallint {u^{ - n}} du = \frac{{{u^{ - n + 1}}}}{{ - n + 1}} + C\) This can also be written as: \(\frac{{{u^{1 - n}}}}{{1 - n}} + C\)
- Substitute back to \(x\): Replace \(u\) with \(1 + \ln x\) in the result: \(\frac{{{{\left( {1 + In\;x} \right)}^{1 - n}}}}{{1 - n}} + C\)
Simplifying and Matching with Options
Let's simplify the result obtained in step 5 to match the given options. We have \(\frac{{{{\left( {1 + In\;x} \right)}^{1 - n}}}}{{1 - n}} + C\).
Using the property \(a^{m-n} = \frac{1}{a^{n-m}}\), we can rewrite \({\left( {1 + In\;x} \right)^{1 - n}}\) as \(\frac{1}{{{\left( {1 + In\;x} \right)^{n - 1}}}}\). So, the expression becomes: \(\frac{{\frac{1}{{{{\left( {1 + In\;x} \right)}^{n - 1}}}}}}{{1 - n}} + C\) This simplifies to: \(\frac{1}{{(1 - n){{\left( {1 + In\;x} \right)}^{n - 1}}}} + C\) Since \(1 - n = -(n - 1)\), we can write this as: \(\frac{1}{{-(n - 1){{\left( {1 + In\;x} \right)}^{n - 1}}}} + C\) or \(- \frac{1}{{(n - 1){{\left( {1 + In\;x} \right)}^{n - 1}}}} + C\) This matches one of the given options.
Conclusion
Comparing our result with the provided options:
- Option 1: \( \frac{1}{{\left( {n - 1} \right){{\left( {1 + In\;x} \right)}^{n - 1}}}} + c \) - Does not match the negative sign.
- Option 2: \( \frac{{1 - n}}{{{{\left( {1 + In\;x} \right)}^{1 - n}}}} + c \) - Does not match the form.
- Option 3: \( \frac{{n + 1}}{{{{\left( {1 + In\;x} \right)}^{n + 1}}}} + c \) - Does not match the form or power.
- Option 4: \( - \frac{1}{{\left( {n - 1} \right){{\left( {1 + In\;x} \right)}^{n - 1}}}} + c \) - Matches our derived result.
Therefore, the correct answer is \( - \frac{1}{{\left( {n - 1} \right){{\left( {1 + In\;x} \right)}^{n - 1}}}} + c \).
| Step | Action | Expression |
|---|---|---|
| 1 | Original Integral | \( \smallint \frac{{dx}}{{x{{\left( {1 + In\;x} \right)}^n}}} \) |
| 2 | Substitution: \(u = 1 + \ln x\) | \(du = \frac{1}{x} dx\) |
| 3 | Integral in terms of \(u\) | \( \smallint u^{-n} du \) |
| 4 | Integrate using Power Rule | \( \frac{u^{1-n}}{1-n} + C \) |
| 5 | Substitute back \(u = 1 + \ln x\) | \( \frac{(1 + \ln x)^{1-n}}{1-n} + C \) |
| 6 | Simplify (as \( \frac{1}{-(n-1)(1+\ln x)^{n-1}} \)) | \( - \frac{1}{{(n - 1){{\left( {1 + In\;x} \right)}^{n - 1}}}} + C \) |
Revision Table: Key Integration Concepts
| Formula | Description |
|---|---|
| \( \smallint x^k dx = \frac{x^{k+1}}{k+1} + C \) (for \(k \ne -1\)) | Power rule for integration. |
| \( \smallint \frac{1}{x} dx = \ln|x| + C \) | Integral of \(1/x\). |
| Substitution Method | A technique used to simplify integrals by replacing variables. Useful when the integrand contains a function and its derivative. |
Additional Information: Understanding Indefinite Integrals
An indefinite integral represents the family of all antiderivatives of a given function. The constant of integration, denoted by \(C\) or \(c\), is added because the derivative of any constant is zero. This means that if \(F(x)\) is an antiderivative of \(f(x)\), then \(F(x) + C\) is also an antiderivative for any constant \(C\). The expression \(\smallint f(x) dx = F(x) + C\) signifies that the derivative of \(F(x) + C\) with respect to \(x\) is \(f(x)\).
The substitution method is a fundamental technique in calculus that reverses the chain rule for differentiation. It helps transform complex integrals into simpler forms that can be solved using standard integration formulas.
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