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Magazine Chapter 8: Problem 41
Evaluate the integral by first using substitution or integration by parts and then using partial fractions. $$ \int x \tan ^{-1} x d x $$
Short Answer
Expert verified
\( \frac{x^2 \tan^{-1}(x)}{2} - \frac{x}{2} + \frac{\tan^{-1}(x)}{2} + C \)
Step by step solution
01
Choose a substitution for easier integration
This problem involves a product of functions, and the inverse tangent function is present. It makes sense to use integration by parts. We can let \( u = \tan^{-1}(x) \) and \( dv = x \, dx \). Consequently, we derive \( du = \frac{1}{1+x^2} \, dx \) and \( v = \frac{x^2}{2} \) through integration.
02
Apply integration by parts formula
The integration by parts formula is \( \int u \, dv = uv - \int v \, du \). Substituting our values, we get:\[ \int x \tan^{-1}(x) \, dx = \tan^{-1}(x) \cdot \frac{x^2}{2} - \int \frac{x^2}{2} \cdot \frac{1}{1+x^2} \, dx \] This simplifies to:\[ \frac{x^2 \tan^{-1}(x)}{2} - \frac{1}{2} \int \frac{x^2}{1+x^2} \, dx \]
03
Simplify the integrand \( \frac{x^2}{1+x^2} \)
Rewriting the integrand, \( \frac{x^2}{1+x^2} \), leads us to consider partial fractions. First, notice that:\[ \frac{x^2}{1+x^2} = 1 - \frac{1}{1+x^2} \]Thus, the integral becomes:\[ \int \frac{x^2}{1+x^2} \, dx = \int 1 \, dx - \int \frac{1}{1+x^2} \, dx \]
04
Integrate using partial fraction decomposition
Each part is straightforward to integrate:\[ \int 1 \, dx = x \] and \[ \int \frac{1}{1+x^2} \, dx = \tan^{-1}(x) \]So, \[ \int \frac{x^2}{1+x^2} \, dx = x - \tan^{-1}(x) \].
05
Substitute back into the integration by parts result
Substitute the result from the partial fractions integral back into the integration by parts formula:\[ \frac{x^2 \tan^{-1}(x)}{2} - \frac{1}{2} (x - \tan^{-1}(x)) \]Simplify to get:\[ \frac{x^2 \tan^{-1}(x)}{2} - \frac{x}{2} + \frac{\tan^{-1}(x)}{2} + C \]where \( C \) is the constant of integration.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Integration by Parts
Integration by parts is a technique used to find the integral of the product of two functions. It resembles the product rule of differentiation but is applied to integration.
Here鈥檚 how it works:
Here鈥檚 how it works:
- Select two parts from the integrand: one to differentiate and the other to integrate. Usually, the inverse function or log function is chosen for differentiation, as they get simpler upon differentiation.
- Write the integral in the form \( \int u \, dv \) where \( u \) is one part and \( dv \) is the differential of the other part.
- Derive \( du \) by differentiating \( u \), and find \( v \) by integrating \( dv \).
- Substitute into the formula \( \int u \, dv = uv - \int v \, du \) to evaluate the integral.
Partial Fraction Decomposition
Partial fraction decomposition is a method used to break complex rational functions into simpler fractions, making them easier to integrate. This is most useful when dealing with integrands that can be expressed as a quotient where the numerator degree is less than the denominator.
Here's a simple approach:
Here's a simple approach:
- Ensure the numerator's degree is less than the denominator's. If needed, perform polynomial long division first.
- Factor the denominator completely if possible. This reveals linear or irreducible quadratic factors to decompose the fraction.
- Express the function as a sum of partial fractions. Solve for unknown coefficients using algebra, by substituting suitable values or equating coefficients.
Inverse Trigonometric Functions
Inverse trigonometric functions, like \( \tan^{-1}(x) \), play a crucial role in many integrations. These functions help unravel angles from given trigonometric values, with their derivatives often appearing in integrals.
Understanding these basics can help:
Understanding these basics can help:
- Know the derivatives: for instance, \( \frac{d}{dx}(\tan^{-1}(x)) = \frac{1}{1+x^2} \), which was useful in our integration by parts step.
- Recognize their appearance in integrands, often simplifying calculations or setting up integration techniques like substitution or by parts.
- Utilize these functions鈥� property of reducing complex functions into simpler form after elaborations like partial fraction decomposition.
