Solving the Integral \(\smallint \frac{{{\rm{dx}}}}{{{\rm{x}}\left( {{{\rm{x}}^7} + 1} \right)}}\)

The problem asks us to evaluate the indefinite integral \(\smallint \frac{{{\rm{dx}}}}{{{\rm{x}}\left( {{{\rm{x}}^7} + 1} \right)}}\). This is a common type of integral involving rational functions, which can often be simplified using substitution or partial fraction decomposition.

Method: Substitution and Partial Fractions

Let's analyze the structure of the integrand. We have an \(x\) term and an \((x^7+1)\) term in the denominator. A useful technique for integrals involving terms like \((x^n + 1)\) is to manipulate the expression to allow for a substitution based on the power of \(x\).

We can multiply the numerator and the denominator by \(x^6\). This will give us \(x^7\) in the denominator multiplying \((x^7+1)\) and \(x^6 \, dx\) in the numerator, which is related to the derivative of \(x^7\).

The integral becomes:

$$\smallint \frac{{{\rm{dx}}}}{{{\rm{x}}\left( {{{\rm{x}}^7} + 1} \right)}} = \smallint \frac{{{\rm{dx}}} \cdot {{\rm{x}}^6}}{{{\rm{x}}\left( {{{\rm{x}}^7} + 1} \right)} \cdot {{\rm{x}}^6}} = \smallint \frac{{{{\rm{x}}^6}{\rm{dx}}}}{{{{\rm{x}}^7}\left( {{{\rm{x}}^7} + 1} \right)}}$$

Now, we can perform a substitution. Let \(u = x^7\). Then the differential \(du\) is \(du = 7x^6 \, dx\). This means \(x^6 \, dx = \frac{1}{7} du\).

Substituting \(u\) and \(du\) into the integral:

$$\smallint \frac{{{{\rm{x}}^6}{\rm{dx}}}}{{{{\rm{x}}^7}\left( {{{\rm{x}}^7} + 1} \right)}} = \smallint \frac{{\frac{1}{7}{\rm{du}}}}{{{\rm{u}}\left( {{\rm{u}} + 1} \right)}} = \frac{1}{7}\smallint \frac{{{\rm{du}}}}{{{\rm{u}}\left( {{\rm{u}} + 1} \right)}}$$

Now we need to evaluate the integral \(\smallint \frac{{{\rm{du}}}}{{{\rm{u}}\left( {{\rm{u}} + 1} \right)}}\). This integral can be solved using partial fraction decomposition.

We want to find constants A and B such that:

$$\frac{1}{{{\rm{u}}\left( {{\rm{u}} + 1} \right)}} = \frac{{\rm{A}}}{{\rm{u}}} + \frac{{\rm{B}}}{{{\rm{u}} + 1}}$$

Multiply both sides by \(u(u+1)\) to clear the denominators:

$$1 = {\rm{A}}\left( {{\rm{u}} + 1} \right) + {\rm{Bu}}$$

We can find A and B by choosing convenient values for u:

• Set \(u = 0\): \(1 = {\rm{A}}(0 + 1) + {\rm{B}}(0) \implies 1 = {\rm{A}}\). So, \(A = 1\).
• Set \(u = -1\): \(1 = {\rm{A}}(-1 + 1) + {\rm{B}}(-1) \implies 1 = -\rm{B}\). So, \(B = -1\).

Thus, the partial fraction decomposition is:

$$\frac{1}{{{\rm{u}}\left( {{\rm{u}} + 1} \right)}} = \frac{1}{{\rm{u}}} - \frac{1}{{{\rm{u}} + 1}}$$

Now substitute this back into the integral with the \(\frac{1}{7}\) factor:

$$\frac{1}{7}\smallint \left( {\frac{1}{{\rm{u}}} - \frac{1}{{{\rm{u}} + 1}}} \right){\rm{du}}$$

Integrate each term:

$$\smallint \frac{1}{{\rm{u}}}{\rm{du}} = \ln \left| {\rm{u}} \right| + {\rm{C_1}}$$$$\smallint \frac{1}{{{\rm{u}} + 1}}{\rm{du}} = \ln \left| {{\rm{u}} + 1} \right| + {\rm{C_2}}$$

So, the integral in terms of u is:

$$\frac{1}{7}\left( {\ln \left| {\rm{u}} \right| - \ln \left| {{\rm{u}} + 1} \right|} \right) + {\rm{C}}$$

Using the property of logarithms \(\ln a - \ln b = \ln \left( \frac{a}{b} \right)\), we get:

$$\frac{1}{7}\ln \left| {\frac{{\rm{u}}}{{{\rm{u}} + 1}}} \right| + {\rm{C}}$$

Finally, substitute back \(u = x^7\) to express the result in terms of x:

$$\frac{1}{7}\ln \left| {\frac{{{{\rm{x}}^7}}}{{{{\rm{x}}^7} + 1}}} \right| + {\rm{C}}$$

This matches one of the given options.

Summary of Steps

To solve the integral \(\smallint \frac{{{\rm{dx}}}}{{{\rm{x}}\left( {{{\rm{x}}^7} + 1} \right)}}\):

1. Multiply numerator and denominator by \(x^6\) to get \(\smallint \frac{{{{\rm{x}}^6}{\rm{dx}}}}{{{{\rm{x}}^7}\left( {{{\rm{x}}^7} + 1} \right)}}\).
2. Use the substitution \(u = x^7\), which gives \(du = 7x^6 \, dx\), or \(x^6 \, dx = \frac{1}{7} du\).
3. Rewrite the integral in terms of \(u\): \(\frac{1}{7}\smallint \frac{{{\rm{du}}}}{{{\rm{u}}\left( {{\rm{u}} + 1} \right)}}\).
4. Use partial fraction decomposition for \(\frac{1}{{{\rm{u}}\left( {{\rm{u}} + 1} \right)}}\) to get \(\frac{1}{{\rm{u}}} - \frac{1}{{{\rm{u}} + 1}}\).
5. Integrate the decomposed terms: \(\frac{1}{7}\smallint \left( {\frac{1}{{\rm{u}}} - \frac{1}{{{\rm{u}} + 1}}} \right){\rm{du}} = \frac{1}{7}\left( {\ln \left| {\rm{u}} \right| - \ln \left| {{\rm{u}} + 1} \right|} \right) + {\rm{C}}\).
6. Simplify using logarithm properties: \(\frac{1}{7}\ln \left| {\frac{{\rm{u}}}{{{\rm{u}} + 1}}} \right| + {\rm{C}}\).
7. Substitute back \(u = x^7\): \(\frac{1}{7}\ln \left| {\frac{{{{\rm{x}}^7}}}{{{{\rm{x}}^7} + 1}}} \right| + {\rm{C}}\).

Final Answer

The value of the integral \(\smallint \frac{{{\rm{dx}}}}{{{\rm{x}}\left( {{{\rm{x}}^7} + 1} \right)}}\) is \(\frac{1}{7}{\rm{In\;}}\left| {\frac{{{{\rm{x}}^7}}}{{{{\rm{x}}^7} + 1}}} \right| + {\rm{c}}\).

Revision Table: Key Integration Techniques

Additional Information: Rational Function Integrals

Integrals of rational functions (functions that are ratios of polynomials) are a core topic in calculus. The general strategy for integrating \(\smallint \frac{P(x)}{Q(x)} dx\) involves several steps:

• Polynomial Long Division: If the degree of the numerator \(P(x)\) is greater than or equal to the degree of the denominator \(Q(x)\), perform polynomial long division to write \(\frac{P(x)}{Q(x)} = S(x) + \frac{R(x)}{Q(x)}\), where \(S(x)\) is the quotient polynomial and \(\frac{R(x)}{Q(x)}\) is the remainder term with degree of \(R(x)\) less than degree of \(Q(x)\). Integrating \(S(x)\) is straightforward.
• Factor the Denominator: Factor the denominator polynomial \(Q(x)\) into linear and irreducible quadratic factors.
• Partial Fraction Decomposition: Express the proper rational function \(\frac{R(x)}{Q(x)}\) as a sum of simpler fractions based on the factors of \(Q(x)\). Linear factors \((ax+b)^n\) correspond to terms \(\frac{A_1}{ax+b} + \dots + \frac{A_n}{(ax+b)^n}\). Irreducible quadratic factors \((ax^2+bx+c)^n\) correspond to terms \(\frac{B_1x+C_1}{ax^2+bx+c} + \dots + \frac{B_n x+C_n}{(ax^2+bx+c)^n}\).
• Integrate the Partial Fractions: Integrate each of the simpler resulting fractions. These integrals typically involve logarithms (for linear factors) and inverse tangents or logarithms (for quadratic factors).

The specific integral \(\smallint \frac{{{\rm{dx}}}}{{{\rm{x}}\left( {{{\rm{x}}^7} + 1} \right)}}\) is a clever application where manipulation leads to a form suitable for substitution, which then results in a simple rational function that can be solved by partial fractions. Recognizing such patterns is key to solving complex integration problems.