Question

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

Short answer

Answer: The integral of the function is given by: $$\int \frac{1}{(x+1)(x^2+2x+2)^2} dx = \ln|x+1| + \frac{1}{10}(x^2 + 2x + 2 - \ln|x^2 + 2x + 2|) - \frac{3}{10}(x^2 + 2x + 2)^{-1} + C$$

Step-by-step solution

Partial Fraction Decomposition

First, we need to decompose the integral into partial fractions. The function can be written in the form: $$\frac{1}{(x+1)(x^2+2x+2)^2} = \frac{A}{x+1} + \frac{Bx+C}{(x^2+2x+2)} + \frac{Dx+E}{(x^2+2x+2)^2}$$

Clearing the denominators

Next, we'll clear the denominators by multiplying both sides by \((x+1)(x^2+2x+2)^2\): $$1 = A(x^2+2x+2)^2 + (Bx + C)(x+1)(x^2 + 2x + 2) + (Dx + E)(x + 1)$$

Solving for constants A, B, C, D, and E

To find the values of A, B, C, D, and E, we will use several values of x. The easiest approach is to use x = -1: $$1 = A(1)^2 \Rightarrow A=1$$ Now, let's differentiate both sides with respect to x and then plug in x=-1: $$0 = 2(-1)+2(2B(-1)+C)(1) + D$$ $$0 = -2 +2(C-2B) + D$$ $$C - 2B + D = 1$$ We can simplify the equation by differentiating again: $$0 = -2(2B+1)+(D+1)(1)$$ $$0 = -4B - 2 + D + 1$$ $$D = 4B - 1$$ Lastly, differentiate once more: $$0 = 6B+C$$ Now we have a system of three equations: - $$6B+C = 0$$ - $$C-2B+D=1$$ - $$D=4B-1$$ Solving this system, we get: $$B=\frac{1}{10},\ C=\frac{-3}{5},\ D=\frac{-3}{10}$$ So now we can rewrite our integral with these coefficients: $$\int \frac{1}{(x+1)(x^2+2x+2)^2} dx = \int \left(\frac{1}{x+1} + \frac{\frac{1}{10}x-\frac{3}{5}}{(x^2+2x+2)} + \frac{-\frac{3}{10}x}{(x^2+2x+2)^2}\right) dx$$

Integrating each term

Now we integrate each term separately: $$\int \frac{1}{(x+1)(x^2+2x+2)^2} dx = \int \frac{1}{x+1} dx + \int \frac{\frac{1}{10}x-\frac{3}{5}}{(x^2+2x+2)} dx + \int \frac{-\frac{3}{10}x}{(x^2+2x+2)^2} dx$$ The first integral is straightforward: $$\int \frac{1}{x+1} dx = \ln|x+1| + C_1$$ For the second integral, we can use the substitution \(u = x^2 + 2x + 2\): $$\int \frac{\frac{1}{10}x-\frac{3}{5}}{(x^2+2x+2)} dx = \frac{1}{10}\int \frac{u-1}{u} du = \frac{1}{10}\int (1 - \frac{1}{u}) du = \frac{1}{10}(u - \ln|u|) + C_2 = \frac{1}{10}(x^2 + 2x + 2 - \ln|x^2 + 2x + 2|) + C_2$$ For the third integral, we'll use the substitution \(v = x^2 + 2x + 2\) and note that the expression simplifies significantly: $$\int \frac{-\frac{3}{10}x}{(x^2+2x+2)^2} dx = -\frac{3}{10}\int \frac{1}{v^2} dv = -\frac{3}{10}(v^{-1}) + C_3 = -\frac{3}{10}(x^2 + 2x + 2)^{-1} + C_3$$

Combining the results

Combine the results from the individual integrals: $$\int \frac{1}{(x+1)(x^2+2x+2)^2} dx = \ln|x+1| + \frac{1}{10}(x^2 + 2x + 2 - \ln|x^2 + 2x + 2|) - \frac{3}{10}(x^2 + 2x + 2)^{-1} + C$$ And this is the final result: $$\int \frac{1}{(x+1)(x^2+2x+2)^2} dx = \ln|x+1| + \frac{1}{10}(x^2 + 2x + 2 - \ln|x^2 + 2x + 2|) - \frac{3}{10}(x^2 + 2x + 2)^{-1} + C$$

Key concepts

Integral Calculus

Integral calculus is a branch of calculus that deals with finding the integral of a function. The integral represents the area under the curve of a function on a graph. In problems involving integral calculus, our goal is to determine the function that describes the accumulated value, such as area or volume, from a given rate of change.

In this exercise, you are asked to evaluate the integral \[ \int \frac{d x}{(x+1)\left(x^{2}+2 x+2\right)^{2}} \]which may look complex at first glance. However, using partial fraction decomposition simplifies the integration process by breaking it into simpler fractions. Each fraction can then be integrated separately. Integral calculus offers a powerful set of techniques to evaluate integrals, and partial fractions is just one strategy to deal with rational functions.

Remember to add a constant of integration, often written as \( C \), when computing an indefinite integral, as it encompasses all possible vertical shifts of the antiderivative.

Substitution Method

The substitution method in calculus is a technique where we substitute parts of the integral with a new variable, usually to simplify the integration process. This approach is especially useful when dealing with composite functions or more complicated algebraic expressions.

In this problem, one part of the integral involves substitution to make it simpler. Consider the term\[ \int \frac{-\frac{3}{10}x}{(x^2+2x+2)^2} dx. \]Here, by setting \( v = x^2 + 2x + 2 \), the differential \( dv \) becomes easier to work with. This substitution helps transform the problem into a basic form:

• Change variables, simplifying the denominator.
• Compute the integral in terms of \( v \) which becomes more straightforward.

This method is handy for integrals that do not fall nicely into standard forms, enabling a seamless approach to handle seemingly complicated integrals.

System of Equations

Solving integrals using partial fractions often leads to systems of equations. In this method, we express a complex expression as the sum of simpler fractions, which results in needing to find unknown constants.

For the given problem, the decomposition is:\[ \frac{1}{(x+1)(x^2+2x+2)^2} = \frac{A}{x+1} + \frac{Bx+C}{(x^2+2x+2)} + \frac{Dx+E}{(x^2+2x+2)^2} \]After expanding and equating coefficients, you solve a system of equations to determine \( A \), \( B \), \( C \), \( D \), and \( E \):

• By plugging in strategic values for \( x \), like \( x = -1 \), to simplify equations.
• Differentiating and substituting to form additional equations.
• Solve these equations to find unknown values that allow you to proceed with integration.

Understanding this process is pivotal as it aids in reducing the complexity of functions that are difficult to integrate directly.

Differentiation

Differentiation is the process of finding the derivative of a function. While typically used to determine rates of change, in this exercise it helps check our work and solve systems of equations derived from partial fraction decomposition.

When solving for constants, differentiation is used on both sides of the equation to:

• Derive additional equations relating unknowns.
• Simplify expressions that arise from the expansion of terms.

For instance, differentiating the expression obtained after clearing denominators helps verify consistency and solve for unknown coefficients. Moreover, differentiation plays a crucial role when evaluating integrals, as it allows for checks on the accuracy of antiderivatives derived from integration work.