Problem 47 - Calculus Early Transcendentals

Question

Use the approaches discussed in this section to evaluate the following integrals. $$\int \frac{d x}{x^{1 / 2}+x^{3 / 2}}$$

Step-by-Step Solution

Answer

Question: Evaluate the integral $$\int \frac{dx}{x^{1 / 2}+x^{3 / 2}}$$ Answer: The integral evaluates to $$2\tan^{-1}(x^{1/2}) + C$$

Step 1: Variable substitution

Let us set the variable substitution: $$u = x^{1/2}$$ Now we need to differentiate \(u\) with respect to \(x\) to obtain \(du\): $$\frac{du}{dx} = \frac{1}{2}x^{-1/2}$$ From this equation, we can find \(dx\) in terms of \(du\) and \(x\): $$dx = 2x^{1/2}du = 2udu$$

Step 2: Substitute into the integral

Now we substitute \(u\) and \(dx\) into the integral and simplify it: \begin{align*} \int \frac{dx}{x^{1 / 2}+x^{3 / 2}} &= \int \frac{2udu}{u + u^3} \\ &= 2\int \frac{u}{u(1+u^2)} du \\ &= 2 \int \frac{1}{1+u^2} du \end{align*}

Step 3: Evaluate the integral

Now we can evaluate this integral, as it is a well-known one: $$2 \int \frac{1}{1+u^2} du = 2\tan^{-1}(u) + C$$

Step 4: Back substitution

Finally, we need to substitute back \(x\) to find the answer to the original integral: $$2\tan^{-1}(u) + C = 2\tan^{-1}(x^{1/2}) + C$$ So, the final answer to the integral is: $$\int \frac{dx}{x^{1 / 2}+x^{3 / 2}} = 2\tan^{-1}(x^{1/2}) + C$$

Key Concepts

Variable Substitution

When encountering an integral that is not easily solvable in its original form, one useful technique is variable substitution. This involves replacing a part of the integral with a new variable, simplifying the expression in such a way that it becomes more manageable. Often, a substitution is chosen so that its derivative is also present in the integral, allowing for the differential parts to be replaced neatly.

For example, if we have an integral that involves a function of a square root, as in the provided exercise, setting a new variable equal to that square root part can simplify the integrand drastically. After substitution, the integral is often transformed into one of the basic forms that we recognize and know how to solve. It's like translating a difficult phrase in a foreign language into your native tongue to make it easier to understand. The substitution method can turn a perplexing problem into a familiar one, enabling us to solve integrals we might not otherwise be able to.

Indefinite Integrals

An indefinite integral represents a family of functions that are antiderivatives of the original function. When we perform indefinite integration, we are effectively reversing differentiation. What we're looking for is not a single solution, but rather all the possible functions that, when differentiated, give back the original function.

Since differentiation wipes away constant terms (because the derivative of a constant is zero), indefinite integrals always include a '+ C', representing an arbitrary constant. Remember that indefinite integrals give a general form of the solution and do not evaluate the function for specific values, unlike definite integrals, which calculate the area under the curve for a given interval.

Integration Techniques

As you delve deeper into integral calculus, you'll find that there's a toolbox of integration techniques at your disposal, each tailored to deal with specific types of functions and integrals. Besides variable substitution, which is the focus of the exercise, other methods include integration by parts, partial fraction decomposition, and trigonometric integration, to name a few.

The choice of technique often depends on the form of the integrand and experience in recognizing patterns that indicate which method will be the most effective. Sometimes, it might be necessary to use a combination of techniques to find the integral. These methods are critical to solving more complex integrals, turning seemingly impenetrable problems into simpler ones that are more straightforward to integrate.

Antiderivatives

Antiderivatives, also known as primitive functions, are the inverse of derivatives. The process of finding antiderivatives is what we call integration. In essence, every time you calculate an integral, you are searching for an antiderivative of the function.

For some functions, finding an antiderivative is straightforward, especially if they match familiar derivative forms. However, this is not always the case, which is why we have multiple integration techniques to help us determine the antiderivative. It's essential to remember that antiderivatives are not unique because of the constant term '+ C', which represents an infinite number of possible vertical shifts of the function. Integral calculus is robust because it encompasses a wide variety of functions, even if they at first appear unrelated.