Problem 6
Question
Find the integral. $$ \int \frac{x^{4}-1}{x^{2}+1} d x $$
Step-by-Step Solution
Verified Answer
The solution to the integral \( \int \frac{x^{4}-1}{x^{2}+1} \, dx \) is \( \frac{1}{3}x^3 - x - \arctan(x) + C \)
1Step 1: Divide each term separately
Divide each term in the numerator by the denominator. We get: \( \frac{x^4}{x^2 + 1} - \frac{1}{x^2 + 1} \)
2Step 2: Simplify
Simplify the first term to \( x^2 - 1 \). The simplified version of the integral will be: \( \int (x^2 - 1 - \frac{1}{x^2 + 1}) \, dx \)
3Step 3: Integrate
The integrals are calculated separately as:The integral of \( x^2 \) is \( \frac{1}{3}x^3 \), of \( 1 \) is \( x \) and \( \frac{1}{x^2 + 1} \) is \( \arctan(x) \). So, we get: \( \frac{1}{3}x^3 - x - \arctan(x) + C \) where \( C \) is the constant of integration.
Key Concepts
Integration TechniquesPolynomial Long DivisionTrigonometric Integrals
Integration Techniques
Integration techniques are crucial in calculus to find functions given their derivative, commonly known as integrals. They help us calculate the area under a curve, solve differential equations, and much more. In this problem, we need to find the integral of a rational function: \[ \int \frac{x^{4}-1}{x^{2}+1} \, dx \]Various integration techniques can be applied, such as:
- Substitution: Useful when the integrand features a function and its derivative.
- Partial Fractions: Effective when dealing with rational functions with factors in the denominator.
- By Parts: This technique is useful when the integrand is a product of two functions.
Polynomial Long Division
Polynomial long division is similar to long division with numbers and is often used to simplify integrals involving polynomials. It helps by dividing the degree of the polynomial in the numerator by the degree of the polynomial in the denominator. In our exercise, we performed the polynomial division to rewrite:\[ \frac{x^{4}-1}{x^{2}+1} \]This was done by dividing each term separately, resulting in:\[ \frac{x^4}{x^2 + 1} - \frac{1}{x^2 + 1} \]After simplification, the rational term turned into a simpler polynomial expression, which was more straightforward to integrate: \[ x^2 - 1 - \frac{1}{x^2 + 1} \]This division method is handy when you need to deal with high-degree polynomials or decompose complex rational expressions into simpler terms. It makes the integration process much more manageable.
Trigonometric Integrals
Trigonometric integrals often involve trigonometric functions in the integrand, and specific methods make their integration more straightforward. In exercises like our present case, we encounter the inverse trigonometric functions for integration.In our example, we had a term:\[ \frac{1}{x^2 + 1} \]The integral of this specific function is known to be:\[ \arctan(x) \]This is because the derivative of \( \arctan(x) \) is \( \frac{1}{x^2 + 1} \). The familiarity with inverse trigonometric integrals like this one is essential because similar forms appear frequently within calculus problems.Understanding how to handle these trigonometric integrals aids in recognizing situations where substitution or recognition of standard results can simplify your work significantly. It's fundamental to learn common trigonometric integration results to speed up problem-solving across different calculus problems.
Other exercises in this chapter
Problem 5
Find the general solution of the differential equation and check the result by differentiation. $$ \frac{d y}{d x}=x^{3 / 2} $$
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Evaluate the function. If the value is not a rational number, round your answer to three decimal places. (a) \(\operatorname{csch}^{-1} 2\) (b) \(\operatorname{
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In Exercises 3-8, evaluate the definite integral by the limit definition. $$ \int_{1}^{3} 3 x^{2} d x $$
View solution Problem 6
Evaluate the definite integral of the algebraic function. Use a graphing utility to verify your result. $$ \int_{2}^{7} 3 d v $$
View solution