Problem 43 - Calculus Early Transcendentals

Question

Evaluate the following integrals. $$\int \frac{x^{2}+x+2}{(x+1)\left(x^{2}+1\right)} d x$$

Step-by-Step Solution

Answer

Question: Evaluate the integral $$\int \frac{x^{2}+x+2}{(x+1)(x^{2}+1)} dx$$. Answer: $$\int \frac{x^{2}+x+2}{(x+1)(x^{2}+1)} dx = \ln\left|x+1\right| + \arctan(x) + C$$

Step 1: Perform partial fraction decomposition

We need to decompose the given fraction into simpler fractions. Our goal is to rewrite the given expression as: $$\frac{x^{2} + x + 2}{(x+1)(x^{2}+1)} = \frac{A}{x+1} + \frac{Bx + C}{x^{2}+1}$$ Multiplying both sides by the denominator, we get: $$x^{2}+x+2= A(x^{2}+1) + (Bx + C)(x+1)$$

Step 2: Determine the constants A, B, and C

Now we need to find the values of A, B, and C that will make both sides of the equation equal. Expand the right-hand side of the equation, we have: $$x^{2}+x+2=(A+B) x^{2}+(C+B) x+(A+C)$$ Comparing the coefficients on both sides, we find the following equations: $$A+B=1$$ $$C+B=1$$ $$A+C=2$$ We can solve this system of linear equations to find the constants A, B, and C. Subtracting the second equation from the first equation, we get: $$A-C=0 \Rightarrow A=C$$ Substituting this into the third equation, we get: $$2A=2 \Rightarrow A=1$$ And so, $$C=1$$ Substituting A back into the first equation, we get: $$1+B=1 \Rightarrow B=0$$ Thus, we have found out the constants A, B, and C: $$A=1, B=0, C=1$$ Plugging these constants into our simpler fractions, we get: $$\frac{1}{x+1} + \frac{1}{x^{2}+1}$$

Step 3: Integrate the simpler fractions

Now, we can integrate each fraction individually: $$\int \frac{x^{2}+x+2}{(x+1)(x^{2}+1)} d x = \int \left(\frac{1}{x+1} + \frac{1}{x^{2}+1}\right) d x = \int \frac{1}{x+1} d x + \int \frac{1}{x^{2}+1} d x$$ Integrating the first fraction, we get: $$\int \frac{1}{x+1} d x = \ln\left|x+1\right|+C_1$$ For the second fraction, we recognize that it is the integral of the function \(\arctan(x)\), so: $$\int \frac{1}{x^{2}+1} d x = \arctan(x)+C_2$$

Step 4: Combine the results and write the final answer

Finally, we combine the results of the integration and write the final answer: $$\int \frac{x^{2}+x+2}{(x+1)(x^{2}+1)} d x = \ln\left|x+1\right| + \arctan(x) + C$$ Where $$C=C_1+C_2$$ is a constant of integration.

Key Concepts

Partial Fraction Decomposition

Partial fraction decomposition is an essential technique in calculus for breaking down complex rational functions into simpler fractions. This method is particularly useful when you need to integrate functions like \( \frac{x^{2}+x+2}{(x+1)(x^{2}+1)} \).
It helps transform a complicated expression into a sum of easier-to-integrate terms.

Here's the general idea:

• Express the complex fraction as a sum of simple fractions using unknown constants.
• Multiply through by the denominator to eliminate fractions on both sides.
• Expand and compare coefficients to set up a system of equations.
• Solve this system to find the constants.

For example, we decompose \( \frac{x^{2}+x+2}{(x+1)(x^{2}+1)} \) into \( \frac{A}{x+1} + \frac{Bx + C}{x^{2}+1} \). This step makes integration straightforward by simplifying the expression into parts that are easier to handle independently.

Integration Techniques

Integration techniques vary depending on the form of the integrand, and choosing the right technique can simplify the process significantly. In our exercise, once the partial fractions are found, each fraction can be tackled individually.

Consider the simpler fractions from decomposition:

• For \( \frac{1}{x+1} \), use the natural logarithm: \( \int \frac{1}{x+1} \, dx = \ln|x+1| + C_1 \).
• For \( \frac{1}{x^{2}+1} \), recognize the arctangent function: \( \int \frac{1}{x^{2}+1} \, dx = \arctan(x) + C_2 \).

Understanding and identifying these integral forms are key. For instance, knowing that \( \int \frac{1}{x^{2}+1} \, dx \) is connected to \( \arctan(x) \) can save a lot of effort, and familiarity with basic integrals is incredibly helpful in this context.

Definite and Indefinite Integrals

Integration produces two types of results: definite and indefinite integrals. The focus in our exercise is on indefinite integrals which include an arbitrary constant \( C \).
Indefinite integrals give a family of functions as their solution:

• They represent anti-derivatives of the integrand.
• Include the constant of integration \( C \) because integration can differ by a constant.

The final result of our integration: \( \ln|x+1| + \arctan(x) + C \), illustrates this concept. Here, \( C \) combines the constant terms from each part of the integral. In contrast, definite integrals compute a specific area under a curve over an interval and provide a numerical result without including \( C \).