Problem 33 - Calculus: Graphical, Numerical, Algebraic

Question

In Exercises \(25-46,\) use substitution to evaluate the integral. $$\int \frac{\ln ^{6} x}{x} d x$$

Step-by-Step Solution

Answer

The evaluated integral of the function \(\frac{\ln^{6}x}{x}\) with respect to \(x\) is \(\frac{1}{7}(\ln x)^{7} + C\).

Step 1: Identify an appropriate substitution

Take \(u=\ln x\). Why this substitution is an adequate choice? It's because the derivative of \(\ln x\) is \(\frac{1}{x}\), which is present in the integral. So, when we substitute, the \(\frac{1}{x}\) in the denominator will cancel out.

Step 2: Calculate the differential du

The differential \(du\) is the derivative of \(u\) with respect to \(x\), multiplied by \(dx\). Calculate \(du\) using the substitution \(u = \ln x\), which yields \(du = \frac{1}{x}\ dx\).

Step 3: Substitute into the Integral

Substitute \(u\) and \(du\) into the original integral. The integral then becomes \(\int u^{6} du\). This is a much simpler integral to compute, as it involves a basic power rule.

Step 4: Evaluate the Integral

The integral \(\int u^{6} du\) can be easily evaluated using the Power Rule for Integrals. The Power Rule states that the integral of \(u^n\) with respect to \(u\) is \(\frac{1}{n+1} u^{n+1}\) (plus a constant). Applying this rule, we get \(\frac{1}{7}u^{7} + C\).

Step 5: Substitute Back for \(x\)

Substitute back in for \(x\) using the original substitution \(u = \ln x\). The final result is then \(\frac{1}{7}(\ln x)^{7} + C\).

Key Concepts

Power Rule for Integrals

The Power Rule for Integrals is a straightforward tool for finding the integral of a polynomial expression. When faced with an expression like \(u^n\), the power rule helps in determining the antiderivative without much hassle.
Essentially, if you encounter \(u^n\), where \(u\) is a variable and \(n\) is a real number not equal to -1, its integral is given by \(\int u^n\, du = \frac{1}{n+1} u^{n+1} + C\). Here, \(C\) represents the constant of integration, which is crucial in indefinite integration.

• For example, if \(n = 6\) as seen in the exercise, the integration becomes \(\int u^6\, du = \frac{1}{7} u^7 + C\).
• This process simplifies integration, especially when handling polynomial terms.

By capitalizing on the power rule, evaluating integrals becomes a matter of adjusting the exponent and including the integration constant, making it an indispensable technique in calculus.

Definite and Indefinite Integrals

In calculus, integrals can be either definite or indefinite, serving different purposes. An indefinite integral, like the one in the exercise, finds the general antiderivative of a function.
It is represented by \(\int f(x) \, dx = F(x) + C\), where \(F(x)\) is the antiderivative and \(C\) is the integration constant that accommodates any shifts in the vertical position of the curve. Indefinite integrals don't have boundaries, hence they produce a family of functions.

• This gives flexibility in solving problems where boundaries aren't defined.
• The result \(\frac{1}{7} (\ln x)^7 + C\) from the exercise case is an example of an indefinite integral.

In contrast, definite integrals calculate the exact area under the curve within a defined interval \([a, b]\). Here, the integration result is \(F(b) - F(a)\), producing a numerical value instead of a general function.

Natural Logarithm Integration

Integrating functions involving the natural logarithm \(\ln(x)\) can often involve using substitution. This strategy is vital since the derivative of \(\ln x\) is \(\frac{1}{x}\), a recurring element that often appears in these integrals.
For instance, in the given exercise, choosing \(u = \ln x\) allows the substitution process to simplify the integral to \(\int u^6 du\). This transition from \(x\) to \(u\) simplifies computation significantly.

• The resulting integral becomes manageable, employing familiar rules like the Power Rule for an easier solution.
• Substituting back \(u = \ln x\) at the end reintroduces the original variable in the solution, ensuring accuracy.

This method strengthens understanding of integrating logarithmic functions and highlights the importance of substitution in breaking down complicated integrals into simpler parts.