What is \(\smallint \frac{{{{\rm{x}}^{{\rm{e}} - 1}} + {{\rm{e}}^{{\rm{x}} - 1}}}}{{{{\rm{x}}^{\rm{e}}} + {{\rm{e}}^{\rm{x}}}}}{\rm{dx}}\) equal to?
Finding the Integral of a Rational Function
We are asked to evaluate the indefinite integral: \(\smallint \frac{{{{\rm{x}}^{{\rm{e}} - 1}} + {{\rm{e}}^{{\rm{x}} - 1}}}}{{{{\rm{x}}^{\rm{e}}} + {{\rm{e}}^{\rm{x}}}}}{\rm{dx}}\).
This integral has the form of a fraction where the numerator looks related to the derivative of the denominator. This suggests using the substitution method.
Applying the Substitution Method for Integration
Let the denominator be our substitution variable, \(u\). Let \(u = {{\rm{x}}^{\rm{e}}} + {{\rm{e}}^{\rm{x}}}\).
Now, we need to find the differential \(du\). We differentiate \(u\) with respect to \(x\): \(\frac{{du}}{{dx}} = \frac{d}{{dx}}\left( {{{\rm{x}}^{\rm{e}}} + {{\rm{e}}^{\rm{x}}}} \right)\)
Using the power rule for differentiation (\(\frac{d}{{dx}}\left( {{x^n}} \right) = n{x^{n - 1}}\)) for the term \({{\rm{x}}^{\rm{e}}}\) (where \(e\) is a constant) and the rule for differentiating the exponential function (\(\frac{d}{{dx}}\left( {{e^x}} \right) = {e^x}\)) for the term \({{\rm{e}}^{\rm{x}}}\), we get:
\(\frac{d}{{dx}}\left( {{{\rm{x}}^{\rm{e}}}} \right) = {\rm{e}} \cdot {{\rm{x}}^{{\rm{e}} - 1}}\)
\(\frac{d}{{dx}}\left( {{{\rm{e}}^{\rm{x}}}} \right) = {{\rm{e}}^{\rm{x}}}\)
So, the derivative of \(u\) with respect to \(x\) is:
\(\frac{{du}}{{dx}} = {\rm{e}}{{\rm{x}}^{{\rm{e}} - 1}} + {{\rm{e}}^{\rm{x}}}\)
Now we can write the differential \(du\) as:
\(du = \left( {{\rm{e}}{{\rm{x}}^{{\rm{e}} - 1}} + {{\rm{e}}^{\rm{x}}}} \right){\rm{dx}}\)
Relating du to the Numerator
We observe that the expression inside the parenthesis \(\left( {{\rm{e}}{{\rm{x}}^{{\rm{e}} - 1}} + {{\rm{e}}^{\rm{x}}}} \right)\) is related to the numerator of our original integral, which is \({{{\rm{x}}^{{\rm{e}} - 1}} + {{\rm{e}}^{{\rm{x}} - 1}}}\).
Let's factor out \(e\) from the differential \(du\):
\(du = e\left( {{{\rm{x}}^{{\rm{e}} - 1}} + \frac{{{{\rm{e}}^{\rm{x}}}}}{e}} \right){\rm{dx}}\)
Using the property of exponents \(\frac{{a^m}}{{a^n}} = {a^{m-n}}\), we have \(\frac{{{{\rm{e}}^{\rm{x}}}}}{e} = \frac{{{{\rm{e}}^{\rm{x}}}}}{{{{\rm{e}}^{\rm{1}}}}} = {{\rm{e}}^{{\rm{x}} - 1}}\).
So, the factored differential is:
\(du = e\left( {{{\rm{x}}^{{\rm{e}} - 1}} + {{\rm{e}}^{{\rm{x}} - 1}}} \right){\rm{dx}}\)
Now, we can see that the term \(\left( {{{\rm{x}}^{{\rm{e}} - 1}} + {{\rm{e}}^{{\rm{x}} - 1}}} \right){\rm{dx}}\) which appears in the numerator of our original integral is equal to \(\frac{1}{e}du\).
\({{{\rm{x}}^{{\rm{e}} - 1}} + {{\rm{e}}^{{\rm{x}} - 1}}}{\rm{dx}} = \frac{1}{e}du\)
Rewriting and Evaluating the Integral
Now substitute \(u = {{\rm{x}}^{\rm{e}}} + {{\rm{e}}^{\rm{x}}}\) and \({{{\rm{x}}^{{\rm{e}} - 1}} + {{\rm{e}}^{{\rm{x}} - 1}}}{\rm{dx}} = \frac{1}{e}du\) into the original integral:
\(\smallint \frac{{{{\rm{x}}^{{\rm{e}} - 1}} + {{\rm{e}}^{{\rm{x}} - 1}}}}{{{{\rm{x}}^{\rm{e}}} + {{\rm{e}}^{\rm{x}}}}}{\rm{dx}} = \smallint \frac{{\frac{1}{e}du}}{u}\)
We can pull the constant factor \(\frac{1}{e}\) out of the integral:
\(= \frac{1}{e}\smallint \frac{1}{u}du\)
The integral of \(\frac{1}{u}\) with respect to \(u\) is a standard integral, which is \(\ln|u|\). Don't forget the constant of integration, \(c\).
\(= \frac{1}{e}\ln\left| u \right| + c\)
Finally, substitute back \(u = {{\rm{x}}^{\rm{e}}} + {{\rm{e}}^{\rm{x}}}\):
\(= \frac{1}{e}\ln\left| {{{\rm{x}}^{\rm{e}}} + {{\rm{e}}^{\rm{x}}}} \right| + c\)
Since \(e\) is a positive constant (\(e \approx 2.718\)), \({{\rm{e}}^{\rm{x}}}\) is always positive for real \(x\). For \({{\rm{x}}^{\rm{e}}}\), if we consider real values and standard domain for \(x^e\), the sum \({{\rm{x}}^{\rm{e}}} + {{\rm{e}}^{\rm{x}}}\) is generally positive in typical calculus contexts where such integrals are encountered. Thus, the absolute value can often be omitted.
The result is \(\frac{1}{e}\ln\left( {{{\rm{x}}^{\rm{e}}} + {{\rm{e}}^{\rm{x}}}} \right) + c\).
Comparing with Options
Let's look at the given options:
- \(\frac{{{x^2}}}{2} + c\)
- In (x + e) + c
- In (x e+ e x) + c
- \(\frac{1}{{\rm{e}}}{\rm{\;In\;}}\left( {{{\rm{x}}^{\rm{e}}} + {{\rm{e}}^{\rm{x}}}} \right) + {\rm{c}}\)
Our calculated result matches option 4.
| Step | Description | Expression |
|---|---|---|
| 1 | Identify the integral | \(\smallint \frac{{{{\rm{x}}^{{\rm{e}} - 1}} + {{\rm{e}}^{{\rm{x}} - 1}}}}{{{{\rm{x}}^{\rm{e}}} + {{\rm{e}}^{\rm{x}}}}}{\rm{dx}}\) |
| 2 | Choose substitution \(u\) | \(u = {{\rm{x}}^{\rm{e}}} + {{\rm{e}}^{\rm{x}}}\) |
| 3 | Calculate \(du\) | \(du = \left( {{\rm{e}}{{\rm{x}}^{{\rm{e}} - 1}} + {{\rm{e}}^{\rm{x}}}} \right){\rm{dx}}\) |
| 4 | Relate numerator dx to \(du\) | \({{{\rm{x}}^{{\rm{e}} - 1}} + {{\rm{e}}^{{\rm{x}} - 1}}}{\rm{dx}} = \frac{1}{e}du\) |
| 5 | Rewrite integral in terms of \(u\) | \(\smallint \frac{1}{u} \cdot \frac{1}{e}du\) |
| 6 | Evaluate the integral | \(\frac{1}{e}\ln\left| u \right| + c\) |
| 7 | Substitute back \(u\) | \(\frac{1}{e}\ln\left| {{{\rm{x}}^{\rm{e}}} + {{\rm{e}}^{\rm{x}}}} \right| + c\) |
| 8 | Final answer (omitting absolute value) | \(\frac{1}{e}\ln\left( {{{\rm{x}}^{\rm{e}}} + {{\rm{e}}^{\rm{x}}}} \right) + c\) |
Revision Table: Key Concepts for Integration
| Concept | Description | Example/Formula |
|---|---|---|
| Indefinite Integral | The reverse process of differentiation. Represents the family of functions whose derivative is the integrand. | \(\smallint f(x)dx = F(x) + c\) if \(F'(x) = f(x)\) |
| Substitution Method | A technique to simplify integrals by replacing the variable of integration with a new variable and its differential. Useful when the integrand contains a function and its derivative (or a constant multiple of it). | \(\smallint f(g(x))g'(x)dx = \smallint f(u)du\) where \(u=g(x)\) and \(du=g'(x)dx\) |
| Derivative of \(x^n\) | Power rule for differentiation. | \(\frac{d}{{dx}}\left( {{x^n}} \right) = n{x^{n - 1}}\) |
| Derivative of \(e^x\) | Derivative of the natural exponential function. | \(\frac{d}{{dx}}\left( {{e^x}} \right) = {e^x}\) |
| Integral of \(\frac{1}{u}\) | Standard integral. | \(\smallint \frac{1}{u}du = \ln\left| u \right| + c\) |
Additional Information: Calculus Integration Techniques
Besides the substitution method used here, there are several other important techniques for evaluating integrals in calculus:
- Integration by Parts: Used for integrating products of functions. The formula is \(\smallint u dv = uv - \smallint v du\). Choosing \(u\) and \(dv\) correctly is key.
- Trigonometric Integrals: Techniques for integrals involving trigonometric functions, often using identities to simplify the integrand.
- Trigonometric Substitution: Used for integrals involving expressions like \(\sqrt{a^2 - x^2}\), \(\sqrt{a^2 + x^2}\), or \(\sqrt{x^2 - a^2}\) by substituting trigonometric functions for the variable \(x\).
- Partial Fraction Decomposition: Used for integrating rational functions (polynomial divided by polynomial) by breaking the fraction down into simpler fractions.
- Improper Integrals: Integrals where the interval of integration is infinite or the integrand has a discontinuity within the interval. These are evaluated using limits.
Understanding when to apply each method is crucial for solving various integration problems. The substitution method is particularly useful when you can identify a function and its derivative within the integrand, as seen in this problem involving \({{\rm{x}}^{\rm{e}}} + {{\rm{e}}^{\rm{x}}}\) and its derivative's scaled version.
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