Problem 39 - Calculus Early Transcendentals

Question

Evaluate the following integrals. $$\int \frac{x^{2}}{\left(25+x^{2}\right)^{2}} d x$$

Step-by-Step Solution

Answer

Question: Evaluate the integral, $$\int \frac{x^2}{\left(25+x^2\right)^2} dx$$. Answer: $$\int \frac{x^{2}}{\left(25+x^{2}\right)^{2}} d x = \frac{1}{10}\arctan\left(\frac{x}{5}\right) - \frac{x}{100(x^2 + 25)} + C$$

Step 1: Identify the Substitution

For this integral, we see that the denominator has the form \((25 + x^2)^2\). A good substitution choice would be to let \(u = 25 + x^2\).

Step 2: Differentiate the Substitution

To proceed, we need to calculate \(du\) in terms of \(dx\). Differentiate \(u\) with respect to \(x\): $$\frac{d u}{d x} = \frac{d}{d x}(25 + x^2) = 2x$$ Now, solve for \(dx\): $$d x=\frac{d u}{2x}$$

Step 3: Substituting \(u\) and \(du\) into the Integral

Replace \(x^2\) and \(dx\) in the original integral with the terms involving \(u\): $$\int \frac{x^2}{\left(25+x^{2}\right)^{2}} d x = \int \frac{u - 25}{u^2} \cdot \frac{du}{2x}$$ Observe that \(x^2 = u - 25\), so we have: $$x = \sqrt{u-25}$$ Substitute this back into the integral: $$\int \frac{u - 25}{u^2} \cdot \frac{du}{2\sqrt{u-25}} = \frac{1}{2} \int \frac{u - 25}{u^2 \sqrt{u-25}} du$$

Step 4: Separate the Terms in the Integrals

Split the integrand into two separate fractions: $$\frac{1}{2} \int \frac{u - 25}{u^2 \sqrt{u-25}} du = \frac{1}{2} \int \left(\frac{1}{u\sqrt{u-25}} - \frac{25}{u^2\sqrt{u-25}}\right) du$$

Step 5: Evaluate the First Integral Separately

Evaluate the first integral: $$\int \frac{1}{u\sqrt{u-25}} du$$ Consider another substitution \(v = \sqrt{u - 25}\) which gives \(v^2 = u - 25\). Differentiating with respect to \(v\) and solving for \(du\) yields: $$du = 2v dv$$ Now, substituting \(v\) in the integral: $$\int \frac{1}{u\sqrt{u-25}} du = 2 \int \frac{v}{(v^2 + 25)v} dv = 2 \int \frac{1}{v^2 + 25} dv$$ Now, notice that this integral is in the form of \(\int \frac{1}{a^2 + x^2} dx\) with \(a=5\). The result of this integral is given by \(\frac{1}{a}\arctan(\frac{x}{a}) + C = \frac{1}{5}\arctan(\frac{v}{5}) + C_1\). Substituting \(v\) back in terms of \(u\), we obtain: $$\frac{1}{5}\arctan\left(\frac{\sqrt{u - 25}}{5}\right) + C_1$$

Step 6: Evaluate the Second Integral Separately

Evaluate the second integral: $$-25\int \frac{1}{u^2\sqrt{u-25}} du$$ Make the substitution \(w = \sqrt{u - 25}\), which gives \(w^2 = u - 25\). Differentiating with respect to \(w\) and solving for \(du\) yields: $$du = 2w dw$$ Performing the substitution in the integral, we have: $$-25\int \frac{1}{u^2\sqrt{u-25}} du = -25 \int \frac{2w}{(w^2 + 25)^2}dw$$ Now, the integral is of the form \(\int \frac{x}{(a^2 + x^2)^2} dx\) with \(a=5\). The result of this integral is given by \(\frac{1}{2a^2}\left(\frac{x}{a^2 + x^2}\right) + C = \frac{w}{2 \cdot 5^2(w^2 + 25)} + C_2\). Substituting \(w\) back in terms of \(u\) get: $$\frac{\sqrt{u - 25}}{50(u - 25 + 25)} + C_2 = \frac{\sqrt{u-25}}{50u} + C_2$$

Step 7: Combine Both Results and Undo the \(u\) Substitution

Combine both results and apply \(u\)-substitution: $$\frac{1}{2} \left(\frac{1}{5}\arctan\left(\frac{\sqrt{u - 25}}{5}\right) + C_1\right) - \frac{1}{2}\left(\frac{\sqrt{u-25}}{50u} + C_2\right)$$ Replace \(u\) with \(x^2 + 25\): $$\frac{1}{10}\arctan\left(\frac{\sqrt{x^2}}{5}\right) + \frac{1}{2}C_1 - \frac{\sqrt{x^2}}{100(x^2 + 25)} - \frac{1}{2}C_2$$ Finally, considering constants \(C\) such that \(C = \frac{1}{2}C_1 + \frac{1}{2}C_2\), the solution to the given integral is: $$\int \frac{x^{2}}{\left(25+x^{2}\right)^{2}} d x = \frac{1}{10}\arctan\left(\frac{x}{5}\right) - \frac{x}{100(x^2 + 25)} + C$$

Key Concepts

Integration techniques

Understanding integration techniques is key in calculus. These techniques allow us to evaluate integrals that are not straightforward. One popular technique is substitution, which is particularly useful for integrals involving compositions of functions. By identifying a part of the integral that can be set as a single variable (often called a *u*-substitution), we simplify the integral into a more easily solvable form.

• Substitution helps simplify complex integrals by transforming the variable.
• Each technique has specific forms and patterns, helping guide its application.

Integrals can also involve alternative methods like integration by parts, partial fractions, or trigonometric identities. Each method serves to tackle different types of integrands, helping solve what might initially seem like complicated problems.

Trigonometric substitution

Trigonometric substitution is a specific type of substitution used for integrals involving radicals of quadratic expressions. Consider expressions like \(\sqrt{a^2 - x^2}\). Here, trigonometric identities can greatly simplify the integrand by transforming the expression into a known trigonometric form.

• Useful for solving integrals involving square roots of sums and differences.
• Involves substituting trigonometric functions like sine, cosine, or tangent.

For instance, if you see \(\sqrt{25 + x^2}\), using the substitution \(x = 5\tan(\theta)\) is beneficial. Then, the expression simplifies using the identity \(1 + \tan^2(\theta) = \sec^2(\theta)\), making it easier to integrate. Proper trigonometric substitution turns the integral into a manageable trigonometric function.

Indefinite integrals

Indefinite integrals are integrals without specified limits of integration. They represent a family of functions (antiderivatives) F(x) whose derivative is the original function. The result of an indefinite integral is expressed as \(F(x) + C\), where \(C\) is the constant of integration. It's crucial to include \(C\) since differentiating makes it impossible to determine this constant.

• Represents a general solution without specific intervals.
• Each solution has an arbitrary constant indicating infinite possible functions.

For an expression like \(\int f(x) \, dx\), finding the antiderivative involves recognizing the pattern of differentiation and reversing it. Mastering indefinite integrals is essential because they form the basis of solving real-world integration problems.

Substitution method

The substitution method simplifies evaluating integrals by changing the variable. The key idea is to make the integral resemble a form you know how to solve. This method often involves setting part of the integral to \(u\), finding \(du\), and replacing terms to transform the entire integral. It's especially useful when dealing with composite functions.

• Replaces complex integrands with easier expressions.
• Simplifies the integration process through variable change.

For instance, in the original problem, noticing \(x^2 + 25\) in the denominator leads us to set \(u = x^2 + 25\). The derivative \(du = 2x \cdot dx\) helps transform the integral into terms of \(u\), making it feasible. This changes complex integrations into straightforward problems, leveraging our understanding of basic function integration.