Question

Refer to the summary box (Partial Fraction Decompositions) and evaluate the following integrals. $$\int \frac{d x}{(x+1)\left(x^{2}+2 x+2\right)^{2}}$$

Short answer

Based on the given step-by-step solution, create a short answer question: Question: Evaluate the integral $$\int\frac{1}{(x+1)(x^2+2x+2)^2}dx$$ Answer: $$\frac{1}{2}\left[2\arctan(x+1)-\frac{1}{2}\ln((x+1)^2+1)\right]+\frac{\frac{1}{2}x-1}{x^2+2x+2}+C$$

Step-by-step solution

Perform Partial Fraction Decomposition

We need to rewrite the given function as the sum of simpler functions that can be integrated easily. Given function is $$\frac{1}{(x+1)(x^2+2x+2)^2}$$. Let's rewrite it as follows: $$\frac{1}{(x+1)(x^2+2x+2)^2} = \frac{A}{x+1} + \frac{Bx+C}{x^2+2x+2} + \frac{Dx+E}{(x^2+2x+2)^2}$$ Now, let's find coefficients A, B, C, D, and E.

Clear the Denominator and Compare Coefficients

To determine the coefficients A, B, C, D, and E, we need to clear the denominators and equate the coefficients of each power of x. To clear the denominators, multiply both sides by $$(x+1)(x^2+2x+2)^2$$. We get: $$1 = A(x^2+2x+2)^2 + (Bx+C)(x+1)(x^2+2x+2) + (Dx+E)(x+1)$$ Now, let's compare the coefficients of each power of x:

Getting the System of Equations from Coefficients

We need to find the coefficients of x^0, x^1, x^2, x^3, and x^4 terms on the right-hand side and construct a system of equations. Let's expand the right-hand side: $$1 = A(x^4+4x^3+8x^2+8x+4) +(Bx^3 + Bx^2 + (C+2B)x+C) + (Dx^3+Dx^2+Ex+E)$$ Grouping by powers of x, we get: $$1=(A)x^4+(4A+B+D)x^3+(8A+B+C+D)x^2+(8A+2B+E)x+(4A+C+E)$$ Now, let's write a system of equations by comparing the coefficients: $$\begin{cases} A = 0 \\ 4A+B+D = 0 \\ 8A+B+C+D = 0 \\ 8A+2B+E = 0 \\ 4A+C+E = 1 \end{cases}$$

Solving the System of Equations

Solving the system of equations above, we get: $$A=0$$ $$B=-\frac{1}{2}$$ $$C=1$$ $$D=\frac{1}{2}$$ $$E=-1$$ Thus, we have our partial fraction decomposition as: $$\frac{0}{x+1}+\frac{-\frac{1}{2}x+1}{x^2+2x+2}+\frac{\frac{1}{2}x-1}{(x^2+2x+2)^2}$$

Perform Integration

Now, we can integrate each term one by one: $$\int\frac{-\frac{1}{2}x+1}{x^2+2x+2}dx=\int\frac{1}{2}\cdot\frac{-(x+1)+2}{(x+1)^2+1}dx$$ Let's substitute \(u=x+1\) and \(du=dx\). We get: $$\int\frac{1}{2}\cdot\frac{-u+2}{u^2+1}du$$ The integral can now be split into two integrals: $$\frac{1}{2}\int\frac{2-u}{u^2+1}du = \frac{1}{2}\int\frac{2}{u^2+1}du - \frac{1}{2}\int\frac{u}{u^2+1}du$$ Now, let's perform the integration: $$\frac{1}{2}\left( 2 \arctan(u) - \frac{1}{2}\ln(u^2+1) \right)$$ Substitute back \(u=x+1\): $$\frac{1}{2}\left[2\arctan(x+1)-\frac{1}{2}\ln((x+1)^2+1)\right]+C_1$$ Now, let's integrate the second term of our partial fraction decomposition: $$\int\frac{\frac{1}{2}x-1}{(x^2+2x+2)^2}dx$$ We can see that this is a difficult integral to solve directly. However, we know that this integral is the derivative of the second term of our partial fraction decomposition: $$\frac{d}{dx}\left(\frac{\frac{1}{2}x-1}{x^2+2x+2}\right)=\frac{\frac{1}{2}x-1}{(x^2+2x+2)^2}$$ Therefore, the integral of the second term is just the second term of our partial fraction decomposition: $$\int\frac{\frac{1}{2}x-1}{(x^2+2x+2)^2}dx=\frac{\frac{1}{2}x-1}{x^2+2x+2}+C_2$$

Combine the Integrals

We now have both integrals for the two parts of the partial fraction decomposition: $$\int\frac{-\frac{1}{2}x+1}{x^2+2x+2}dx=\frac{1}{2}\left[2\arctan(x+1)-\frac{1}{2}\ln((x+1)^2+1)\right]+C_1$$ $$\int\frac{\frac{1}{2}x-1}{(x^2+2x+2)^2}dx=\frac{\frac{1}{2}x-1}{x^2+2x+2}+C_2$$ Finally, we can add these two integrals together: $$\int\left(\frac{-\frac{1}{2}x+1}{x^2+2x+2}+\frac{\frac{1}{2}x-1}{(x^2+2x+2)^2}\right)dx = \frac{1}{2}\left[2\arctan(x+1)-\frac{1}{2}\ln((x+1)^2+1)\right]+\frac{\frac{1}{2}x-1}{x^2+2x+2}+C$$ Here C is the sum of constants \(C_1\) and \(C_2\). And this is the final answer.

Key concepts

Integral Calculus

Integral calculus is an essential part of mathematics that is concerned with the accumulation of quantities, such as areas under curves or the total accumulation of change over an interval. It reverses the process of differentiation, helping us find functions when given their derivatives. This branch of calculus is divided into two main parts:

• Definite Integrals: These provide a numerical value representing the area under a curve between two specified points. They're commonly used in physics and engineering to calculate accumulated quantities, like distance or work.


• Indefinite Integrals: These represent general solutions of a set of functions, also known as antiderivatives. They add a constant, as there could be many functions with the same derivative.

When evaluating integrals, especially those involving complex rational functions like in the exercise, we often need specific techniques. This is because they can be daunting to solve by straightforward methods.

Rational Functions

A rational function is essentially a fraction made up of two polynomials. Understanding their properties is vital for several areas of calculus and algebra. A general rational function can be expressed as:\[R(x) = \frac{P(x)}{Q(x)}\]where \(P(x)\) and \(Q(x)\) are polynomials, and \(Q(x) eq 0\).

In the context of integration, rational functions might become complex, especially when they involve higher powers in the denominator, as in this exercise. To simplify these integrals, we use a technique called partial fraction decomposition:

• This method involves breaking down a complex rational function into simpler parts that are easier to integrate.


• It typically requires expressing the function as a sum of fractions with linear or quadratic denominators.


• Identifying each fraction correctly enables easier and more straightforward integration processes.

Rational functions are highly prevalent in real-life applications, involving calculations for growth rates, economics, and much more.

Integration Techniques

Integration techniques are the various methods used to solve integrals that seem complex or unsolvable at first glance. Mastering these techniques is crucial in efficiently tackling integral calculus problems. Some key techniques include:

• Substitution: This technique involves a change of variables to simplify an integral, transforming complex expressions into manageable forms. It's particularly useful when dealing with compositions of functions.


• Integration by Parts: Often used when integrals suggest a product of functions, it relies on the formula:\[\int u \, dv = uv - \int v \, du\]This can simplify integrals involving polynomials or exponentials multiplied by trigonometric or logarithmic functions.


• Partial Fraction Decomposition: As shown in the exercise, this method splits a complex rational function into a sum of simpler fractions, making it easier to integrate each part.

When faced with an integral like the one in this exercise, identifying the right technique or combination of techniques is pivotal. Not only does it ease the computational load, but it also provides deeper insights into the behaviors of the functions involved.