Question

\(\int\left[\left(x^{2} d x\right) /\left\\{\left(x^{2}+2\right)\left(x^{2}+3\right)\right\\}\right]=\ldots\) (a) \(\sqrt{3} \tan ^{-1}(\mathrm{x} / \sqrt{3})+\sqrt{2} \tan ^{-1}(\mathrm{x} / \sqrt{2})\) (b) \(\sqrt{3} \tan ^{-1}(\mathrm{x} / \sqrt{3})-\sqrt{2} \tan ^{-1}(\mathrm{x} / \sqrt{2})\) (c) \(\tan ^{-1}(\mathrm{x} / \sqrt{3})+\sqrt{2} \tan ^{-1}(\mathrm{x} / \sqrt{2})\) (d) \(\tan ^{-1}(\mathrm{x} / \sqrt{3})-\sqrt{2} \tan ^{-1}(\mathrm{x} / \sqrt{2})\)

Short answer

The short answer is: \( \int\left[\frac{x^2}{(x^2+2)(x^2+3)}\right]dx = \sqrt{3} \tan ^{-1}\left(\frac{x}{\sqrt{3}}\right)-\sqrt{2} \tan ^{-1}\left(\frac{x}{\sqrt{2}}\right) + C \)

Step-by-step solution

Partial Fraction Decomposition

Rewrite the integrand as the sum of simpler functions using partial fraction decomposition: \[\frac{x^2}{(x^2+2)(x^2+3)} = \frac{A}{x^2+2} + \frac{B}{x^2+3}\] Multiply both sides by the denominator \((x^2+2)(x^2+3)\) to get: \[x^2 = A(x^2+3) + B(x^2+2)\] #Step 2: Find the coefficients A and B#

Determine Coefficients A and B

The equation above should hold true for any value of x. We can choose two values for x and solve the linear system of equations for A and B. Let \(x=0\), then \[0 = 3A + 2B\] Now, let \(x=1\), then \[1 = 3A + 2B\] After solving the system of linear equations, we find that \(A = 1\) and \(B = -1\). #Step 3: Rewrite the integrand and integrate#

Integrate the Functions

Now we can write the integrand with the coefficients we found: \[\frac{x^2}{(x^2+2)(x^2+3)} = \frac{1}{x^2+2} - \frac{1}{x^2+3}\] Now integrate each term separately: \[\int\left[\frac{1}{x^2+2} - \frac{1}{x^2+3}\right]dx = \int\frac{1}{x^2+2}dx - \int\frac{1}{x^2+3}dx\] For the first integral, let \(x=\sqrt{2}\tan(u)\), then \(dx = \sqrt{2}\sec^2(u)du\). We get: \[\int\frac{1}{x^2+2}dx = \int\frac{\sqrt{2}\sec^2(u)}{2\tan^2(u)+2}du = \frac{1}{\sqrt{2}}\int du = \frac{1}{\sqrt{2}}u + C_1 = \frac{1}{\sqrt{2}}\tan^{-1}\left(\frac{x}{\sqrt{2}}\right) + C_1\] For the second integral, let \(x=\sqrt{3}\tan(v)\), then \(dx = \sqrt{3}\sec^2(v)dv\). We get: \[\int\frac{1}{x^2+3}dx = \int\frac{\sqrt{3}\sec^2(v)}{3\tan^2(v)+3}dv = \frac{1}{\sqrt{3}}\int dv = \frac{1}{\sqrt{3}}v + C_2 = \frac{1}{\sqrt{3}}\tan^{-1}\left(\frac{x}{\sqrt{3}}\right) + C_2\] So, the integral is: \[\int\left[\frac{x^2}{(x^2+2)(x^2+3)}\right]dx = \frac{1}{\sqrt{2}}\tan^{-1}\left(\frac{x}{\sqrt{2}}\right) - \frac{1}{\sqrt{3}}\tan^{-1}\left(\frac{x}{\sqrt{3}}\right) + C\] Comparing with the given alternatives, the correct option is: (b) \(\sqrt{3} \tan ^{-1}(\mathrm{x} / \sqrt{3})-\sqrt{2} \tan ^{-1}(\mathrm{x} / \sqrt{2})\).

Key concepts

Indefinite Integration

Indefinite integration, often termed as anti-differentiation, is a mathematical process to find a function whose derivative is given. It plays a pivotal role in calculus and represents the inverse of differentiation. When we perform indefinite integration, we basically look for all the possible antiderivatives of a function. For instance, if you are asked to integrate a function like \(f(x) = x^2\), the indefinite integral is expressed with the integral sign and the differential \(dx\), which symbolizes the variable of integration. The indefinite integral is written as \(\int f(x) \, dx\) and, in this case, \(\int x^2 \, dx\) which evaluates to \(\frac{1}{3}x^3 + C\), where \(C\) is the constant of integration representing an infinite family of functions due to the reverse process of differentiation.

The example in the exercise demonstrates indefinite integration of a more complex function involving inverse trigonometric functions, made easier by breaking down the function into parts through partial fraction decomposition first.

Linear System of Equations

A linear system of equations is a collection of one or more equations involving the same variables. Typically, these equations are written in the form \(ax + by = c\), where \(a\), \(b\), and \(c\) are constants. Solving a linear system means finding the values of the variables that satisfy all of the equations simultaneously. This often involves techniques like substitution, elimination, or matrix operations.

In the context of partial fraction decomposition, which we see in the exercise, coefficients of the decomposed fractions can be determined by solving a linear system. By equating the coefficients on both sides and choosing suitable values for the variable, we obtain equations in these unknown coefficients, solving which gives us the requisite constants for the integration step. This interplay between algebra and calculus exemplifies the beauty of mathematics where one area supplements another to achieve the solution.

Inverse Trigonometric Functions

Inverse trigonometric functions are the inverses of the trigonometric functions and are used to determine the angle when given the ratio of sides in a right triangle. These functions include the inverse sine \(\sin^{-1}\), inverse cosine \(\cos^{-1}\), and inverse tangent \(\tan^{-1}\) among others. While the basic trigonometric functions take an angle and give a ratio, the inverses take a ratio and return the possible angle that would yield that ratio as part of a right-angled triangle.

In the realm of calculus, these are frequently encountered during the process of integration, especially when dealing with integrals that require substitution to yield results that incorporate \(\tan^{-1}\). Like in the given exercise, after an appropriate substitution, the resulting integral is expressed as a function of \(\tan^{-1}\), indicating a relationship between the algebraic expression and the corresponding angle in trigonometry.

Integration Techniques

There are multiple techniques for integrating complex mathematical functions, each suited for particular types of integrands. Among these are substitution, which is used when an integrand is a function of another function; integration by parts, tailored for products of functions; and partial fractions, ideal for rational functions where the denominator can be factored into linear or quadratic terms.

In the given exercise, partial fraction decomposition is used to simplify the integrand into simpler fractions that are more manageable to integrate. After this step, a trigonometric substitution is applied to integrate each term. Other common techniques include trigonometric integrals, integrals involving exponential and logarithmic functions, and improper integrals. Familiarity with a broad array of integration techniques is essential for students to tackle a wide range of problems in calculus.