Question

Evaluate the following integrals. $$\int_{1}^{4} \frac{d x}{x^{2}-2 x+10}$$

Short answer

In summary, after a series of simplifications and substitutions, the definite integral of the given function is $$\int_{1}^{4} \frac{dx}{x^{2}-2x+10} = \frac{\pi}{6\sqrt{3}} - \frac{\pi}{12}.$$

Step-by-step solution

Rewrite the denominator by completing the square

To complete the square with the denominator, we need to see which perfect square trinomial it resembles. In this case, it resembles \((x-1)^2 + 9\), so we can rewrite the integral as: $$\int_{1}^{4} \frac{1}{(x-1)^2 + 9} dx$$

Choose the appropriate trigonometric substitution

We will use the trigonometric substitution \(x-1 = 3\tan(\theta)\), which leads to \(dx=3\sec^2(\theta)d\theta\). We will also need to find the limits in terms of \(\theta\). When \(x=1\), \(x-1=0\) and \(\tan(\theta)=0\), therefore \(\theta=0\). When \(x=4\), \(x-1=3\) and \(\tan(\theta)=1\), meaning \(\theta=\frac{\pi}{4}\).

Substitute into the integral

Now substitute the expressions for \(x-1\) and \(dx\) into the integral: $$\int_{0}^{\frac{\pi}{4}}\frac{3\sec^2(\theta)}{(3\tan(\theta))^2+9}d\theta$$

Simplify the integral

Simplify the integral as follows: $$\int_{0}^{\frac{\pi}{4}}\frac{3\sec^2(\theta)}{9\tan^2(\theta)+9}d\theta = \int_{0}^{\frac{\pi}{4}}\frac{\sec^2(\theta)}{3\tan^2(\theta)+3}d\theta$$ Now, use the identity \(\sec^2(\theta) = 1 + \tan^2(\theta)\) to simplify the integral further: $$\int_{0}^{\frac{\pi}{4}}\frac{1+\tan^2(\theta)}{3\tan^2(\theta)+3}d\theta = \int_{0}^{\frac{\pi}{4}}\frac{1}{3}+\frac{\tan^2(\theta)}{3\tan^2(\theta)+3}d\theta$$

Evaluate the integral

Separate the integral into two parts and evaluate each part: $$\frac{1}{3}\int_{0}^{\frac{\pi}{4}}d\theta+\int_{0}^{\frac{\pi}{4}}\frac{\tan^2(\theta)}{3\tan^2(\theta)+3}d\theta$$ The first integral is straightforward to compute: $$\frac{1}{3}[\theta]_{0}^{\frac{\pi}{4}} = \frac{\pi}{12}$$ The second integral can be simplified as: $$\int_{0}^{\frac{\pi}{4}}\frac{\tan^2(\theta)}{3\tan^2(\theta)+3}d\theta = \int_{0}^{\frac{\pi}{4}}\frac{1}{3+\frac{3}{\tan^2(\theta)}}d\theta$$ Using the identity \(\cot^2(\theta) = \frac{1}{\tan^2(\theta)}\): $$\int_{0}^{\frac{\pi}{4}}\frac{1}{3+\cot^2(\theta)}d\theta$$ Use the substitution: \(u=\cot(\theta)\), which is equivalent to \(du=-\csc^2(\theta)d\theta\). Then, the integral becomes: $$-\int_{u(\frac{\pi}{4})}^{\infty}\frac{1}{3+u^2}du$$ Now, this integral can be evaluated using the arctangent antiderivative: $$-\left[\frac{1}{\sqrt{3}}\arctan\left(\frac{u}{\sqrt{3}}\right)\right]_{\infty}^{1}+\frac{\pi}{12}$$ Finally, we have: $$-\frac{1}{\sqrt{3}}\left[\arctan\left(\frac{1}{\sqrt{3}}\right)-\arctan(\infty)\right]+\frac{\pi}{12}$$

Find the final answer

Putting everything together, the final answer is: $$-\frac{1}{\sqrt{3}}\left[\frac{\pi}{6}-\frac{\pi}{2}\right]+\frac{\pi}{12} = \frac{\pi}{6\sqrt{3}} - \frac{\pi}{12}$$ So, the integral evaluates to: $$\int_{1}^{4} \frac{dx}{x^{2}-2x+10} = \frac{\pi}{6\sqrt{3}} - \frac{\pi}{12}$$

Key concepts

Trigonometric Substitution

Trigonometric substitution is an integration technique used to simplify integrals involving expressions like square roots and trigonometric identities. In this exercise, we needed to replace the term \(x-1\) with a trigonometric function to ease calculations.
By using the substitution \(x-1 = 3\tan(\theta)\), we transformed our variable into a trigonometric one, allowing us to leverage identities like \sec^2(\theta) = 1 + \tan^2(\theta)\.
This makes the problem simpler and fits well with integration formulae. This approach helps in

• Taking control of complex polynomial exponents.
• Handling expressions that resemble Pythagorean identities easily.
• Simplifying the limits of integration by altering them according to the trigonometric function.

Overall, trigonometric substitution is an invaluable tool to deal with integrals that seem unwieldy at first glance.

Completing the Square

Completing the square is a method used to transform quadratic expressions into a perfect square form. This helps make integrals and equations simpler.
In our exercise, we faced the quadratic expression \(x^2 - 2x + 10\). By rewriting it, we arrived at \((x-1)^2 + 9\).
This simplification is crucial because:

• It allows us to apply trigonometric substitutions smoothly.
• Transforms complex denominators into more manageable forms.
  
• Makes use of formulas and equations related to perfect squares.

Understanding and mastering completing the square can save significant time and help in various mathematical problems, not just integrals.

Antiderivatives

Antiderivatives are the reverse processes of derivatives. When finding the antiderivative, we are essentially looking for the original function given its derivative.
During the evaluation of our integral, we used antiderivatives of trigonometric functions such as arctangent.
Recognizing antiderivatives like \arctan(u)\ is important because:

• It simplifies the evaluation of complex integrals after substitution.
  
• Helps in the transition from variable \(\theta\) back to the original variable.
• Plays a key role in finding the indefinite integrals during calculation.

Antiderivatives provide the foundation for definite integral calculations and are essential for accurate integration across various functions.

Integration Techniques

Integration techniques encompass various methods and substitutions used to find the integral of a function. The choice of technique greatly depends on the form and complexity of the integral.
For our integral, different strategies like trigonometric substitution and completing the square were crucial.
Techniques can include:

• Basic substitution, which is swapping variables for simplification.
• Partial fraction decomposition, useful for rational functions.
  
• Technique choice based on form, such as trigonometric identities or algebraic manipulation.

This variety of techniques allows one to approach nearly any integral confidently.
Mastery of these techniques ensures that any integral, no matter how daunting, can be tackled methodically and successfully.