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Magazine Chapter 7: Problem 71
Evaluate \(\int x \tan ^{-1} x d x\) using integration by parts. Simplify the computation of \(\int v d u\) by introducing a constant of integration \(C_{1}=\frac{1}{2}\) when going from \(d v\) to \(v\).
Short Answer
Expert verified
\( \int x \tan^{-1} x \, dx = \frac{x^2}{2} \tan^{-1}x + \tan^{-1}x - \frac{x}{2} + C \)
Step by step solution
01
Identify Parts for Integration by Parts
Integration by parts is given by the formula \( \int u \, dv = uv - \int v \, du \).We choose \( u = \tan^{-1}x \) and \( dv = x \, dx \).
02
Differentiate and Integrate
Differentiate \( u = \tan^{-1}x \) to find \( du \) and integrate \( dv = x \, dx \) to find \( v \):\[ du = \frac{1}{1+x^2} \, dx \]\[ v = \frac{x^2}{2} + C_1 \]Note, \( C_1 = \frac{1}{2} \) is introduced as a constant of integration.
03
Apply Integration by Parts Formula
Substitute \( u \), \( du \), \( v \), and \( dv \) into the integration by parts formula:\[ \int x \tan^{-1}x \, dx = \tan^{-1}x \cdot \left(\frac{x^2}{2} + \frac{1}{2}\right) - \int \left( \frac{x^2}{2} + \frac{1}{2} \right) \cdot \frac{1}{1+x^2} \, dx \]
04
Simplify the Integral
Simplify the integral \( \int \left( \frac{x^2}{2} + \frac{1}{2} \right) \cdot \frac{1}{1+x^2} \, dx \):This becomes:\[ \frac{1}{2} \int \frac{x^2}{1+x^2} \, dx + \frac{1}{2} \int \frac{1}{1+x^2} \, dx\]Recognizing the second integral as \( \frac{1}{2} \tan^{-1}x \) and using partial fraction decomposition or substitution for the first.
05
Solve the Simplified Integrals
The integral \( \int \frac{1}{1+x^2} \, dx \) evaluates directly to \( \tan^{-1}x \). For \( \int \frac{x^2}{1+x^2} \, dx \), rewrite it as \( \int (1 - \frac{1}{1+x^2}) \, dx \), which becomes simpler to evaluate:\[ \int (1 - \frac{1}{1+x^2}) \, dx = x - \tan^{-1}x \].
06
Combine Results
Combine the evaluated integrals back into the parts formula:\[ \int x \tan^{-1}x \, dx = \tan^{-1}x \cdot \left(\frac{x^2}{2} + \frac{1}{2} \right) - \frac{1}{2}(x - \tan^{-1}x) - \frac{1}{2} \tan^{-1}x + C \].Simplify to get the final expression:\[ \frac{x^2}{2} \tan^{-1}x + \frac{1}{2} \tan^{-1}x - \frac{x}{2} + \frac{1}{2} \tan^{-1}x + C \].
07
Simplify Final Expression
Combine like terms in the expression:\[ \frac{x^2}{2} \tan^{-1}x + \tan^{-1}x - \frac{x}{2} + C \].
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Inverse Trigonometric Functions
Inverse trigonometric functions, such as \( \tan^{-1}x \), are crucial in calculus, especially when evaluating certain integrals. These functions help in finding angles when given a trigonometric ratio, providing a bridge between algebra and geometry.
For instance, \( \tan^{-1}x \) or "arctangent of x" returns the angle whose tangent is x. In this problem, the function \( u = \tan^{-1}x \) is a smart choice in integration by parts. This leads to a derivative that simplifies the integral.
For instance, \( \tan^{-1}x \) or "arctangent of x" returns the angle whose tangent is x. In this problem, the function \( u = \tan^{-1}x \) is a smart choice in integration by parts. This leads to a derivative that simplifies the integral.
- Inverse trigonometric functions are unique in that they are essential in inverse operations, correcting or reversing original trigonometric functions.
- Using this function, derivatives become simple as shown by \( du = \frac{1}{1+x^2} \, dx \).
- It assists learners to manage complex algebraic expressions, transforming them into manageable fractions.
Partial Fraction Decomposition
Partial fraction decomposition is a technique where a complex rational expression is broken down into simpler fractions that are easier to integrate. However, in this exercise, an alternative, simpler strategy is used for decomposing \( \frac{x^2}{1+x^2} \).
Rather than traditional partial fraction decomposition, this expression simplifies directly by rewriting it as \( 1 - \frac{1}{1+x^2} \). This reveals two separate terms manageable individually.
Rather than traditional partial fraction decomposition, this expression simplifies directly by rewriting it as \( 1 - \frac{1}{1+x^2} \). This reveals two separate terms manageable individually.
- This technique is powerful in breaking down difficult integrals so that each part can be dealt with through basic integration.
- Rewriting \( \frac{x^2}{1+x^2} \) helps in reducing computational complexity and speeds up solving steps.
Constant of Integration
The constant of integration is a pivotal concept in integration. When you find an indefinite integral, it represents a family of functions, all differing by a constant. In this particular exercise, a constant \( C_1 = \frac{1}{2} \) is introduced correspondingly when moving from \( dv \) to \( v \).
This artificial introduction is exceptionally useful, simplifying the entire integration process by balancing expressions.
This artificial introduction is exceptionally useful, simplifying the entire integration process by balancing expressions.
- While \( C_1 \) seems minor, its presence is crucial for precise evaluation and simplification.
- The final integral typically includes a constant of integration \( C \), ensuring the general form of the antiderivative.
