Integration Techniques Solving ∫(x³-1)/(x³-x) Dx And ∫₋₄⁰ |x+2| Dx

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This article dives into the solutions for two intriguing calculus problems. We will explore the integration of a rational function, specifically ∫x3−1x3−xdx{ \int \frac{x^3 - 1}{x^3 - x} dx }, and the evaluation of a definite integral involving an absolute value function, ∫−40∣x+2∣dx{ \int_{-4}^{0} |x + 2| dx }. These problems are excellent examples of the techniques required in integral calculus, showcasing partial fraction decomposition and handling absolute values within integrals.

(a) Integrating Rational Functions: ∫x3−1x3−xdx{ \int \frac{x^3 - 1}{x^3 - x} dx }

To solve ∫x3−1x3−xdx{ \int \frac{x^3 - 1}{x^3 - x} dx }, we'll employ a method called partial fraction decomposition. This technique is crucial when dealing with rational functions where the degree of the numerator is not less than the degree of the denominator. Here's a step-by-step breakdown:

1. Polynomial Long Division

Our initial step involves polynomial long division since the degree of the numerator (3) is equal to the degree of the denominator (3). This allows us to rewrite the integrand as a sum of a polynomial and a proper rational function (where the degree of the numerator is less than the degree of the denominator).

Dividing x3−1{ x^3 - 1 } by x3−x{ x^3 - x }, we get:

x3−1x3−x=1+x−1x3−x{ \frac{x^3 - 1}{x^3 - x} = 1 + \frac{x - 1}{x^3 - x} }

This simplifies our integral to:

∫x3−1x3−xdx=∫(1+x−1x3−x)dx{ \int \frac{x^3 - 1}{x^3 - x} dx = \int \left(1 + \frac{x - 1}{x^3 - x}\right) dx }

2. Factor the Denominator

The next crucial step in partial fraction decomposition is factoring the denominator of the proper rational function. In our case, the denominator is x3−x{ x^3 - x }. We can factor out an x{ x }:

x3−x=x(x2−1){ x^3 - x = x(x^2 - 1) }

Furthermore, x2−1{ x^2 - 1 } is a difference of squares, which factors into (x−1)(x+1){ (x - 1)(x + 1) }. Therefore, the fully factored denominator is:

x3−x=x(x−1)(x+1){ x^3 - x = x(x - 1)(x + 1) }

3. Partial Fraction Decomposition

Now, we express the rational function x−1x3−x{ \frac{x - 1}{x^3 - x} } as a sum of simpler fractions, each with one of the factors of the denominator as its denominator. This is the core of the partial fraction decomposition method. We write:

x−1x(x−1)(x+1)=Ax+Bx−1+Cx+1{ \frac{x - 1}{x(x - 1)(x + 1)} = \frac{A}{x} + \frac{B}{x - 1} + \frac{C}{x + 1} }

Our goal is to find the constants A, B, and C. To do this, we multiply both sides of the equation by the common denominator, x(x−1)(x+1){ x(x - 1)(x + 1) }:

x−1=A(x−1)(x+1)+Bx(x+1)+Cx(x−1){ x - 1 = A(x - 1)(x + 1) + Bx(x + 1) + Cx(x - 1) }

4. Solving for the Constants

There are a couple of ways to solve for A, B, and C. One method is to expand the right side and equate coefficients of like powers of x{ x }. Another method, which is often quicker, involves strategically choosing values of x{ x } that will eliminate some of the terms. Let's use the latter approach:

  • Let x=0{ x = 0 }: 0−1=A(0−1)(0+1)+B(0)(0+1)+C(0)(0−1){ 0 - 1 = A(0 - 1)(0 + 1) + B(0)(0 + 1) + C(0)(0 - 1) } −1=A(−1)(1){ -1 = A(-1)(1) } A=1{ A = 1 }

  • Let x=1{ x = 1 }: 1−1=A(1−1)(1+1)+B(1)(1+1)+C(1)(1−1){ 1 - 1 = A(1 - 1)(1 + 1) + B(1)(1 + 1) + C(1)(1 - 1) } 0=0+2B+0{ 0 = 0 + 2B + 0 } B=0{ B = 0 }

  • Let x=−1{ x = -1 }: −1−1=A(−1−1)(−1+1)+B(−1)(−1+1)+C(−1)(−1−1){ -1 - 1 = A(-1 - 1)(-1 + 1) + B(-1)(-1 + 1) + C(-1)(-1 - 1) } −2=0+0+2C{ -2 = 0 + 0 + 2C } C=−1{ C = -1 }

Thus, we have found that A=1{ A = 1 }, B=0{ B = 0 }, and C=−1{ C = -1 }.

5. Substitute Back into the Integral

Now we substitute the values of A, B, and C back into our partial fraction decomposition:

x−1x(x−1)(x+1)=1x+0x−1+−1x+1=1x−1x+1{ \frac{x - 1}{x(x - 1)(x + 1)} = \frac{1}{x} + \frac{0}{x - 1} + \frac{-1}{x + 1} = \frac{1}{x} - \frac{1}{x + 1} }

Our original integral now becomes:

∫(1+x−1x3−x)dx=∫(1+1x−1x+1)dx{ \int \left(1 + \frac{x - 1}{x^3 - x}\right) dx = \int \left(1 + \frac{1}{x} - \frac{1}{x + 1}\right) dx }

6. Integrate

The final step is to integrate each term separately. The integral of 1 is x{ x }, the integral of 1x{ \frac{1}{x} } is ln⁡∣x∣{ \ln|x| }, and the integral of 1x+1{ \frac{1}{x + 1} } is ln⁡∣x+1∣{ \ln|x + 1| }. Therefore, the solution is:

∫(1+1x−1x+1)dx=x+ln⁡∣x∣−ln⁡∣x+1∣+C{ \int \left(1 + \frac{1}{x} - \frac{1}{x + 1}\right) dx = x + \ln|x| - \ln|x + 1| + C }

We can further simplify this using the properties of logarithms:

x+ln⁡∣x∣−ln⁡∣x+1∣+C=x+ln⁡∣xx+1∣+C{ x + \ln|x| - \ln|x + 1| + C = x + \ln\left|\frac{x}{x + 1}\right| + C }

Therefore, the final answer is:

∫x3−1x3−xdx=x+ln⁡∣xx+1∣+C{ \int \frac{x^3 - 1}{x^3 - x} dx = x + \ln\left|\frac{x}{x + 1}\right| + C }

(b) Evaluating Definite Integrals with Absolute Values: ∫−40∣x+2∣dx{ \int_{-4}^{0} |x + 2| dx }

Now, let's tackle the problem of evaluating the definite integral ∫−40∣x+2∣dx{ \int_{-4}^{0} |x + 2| dx }. The presence of the absolute value function, ∣x+2∣{ |x + 2| }, requires us to consider different cases based on the sign of the expression inside the absolute value.

1. Understanding Absolute Value

The absolute value of a number is its distance from zero. This means ∣x∣=x{ |x| = x } if x≥0{ x \geq 0 } and ∣x∣=−x{ |x| = -x } if x<0{ x < 0 }. Applying this to our case, we have:

∣x+2∣={x+2if x+2≥0 (i.e., x≥−2)−(x+2)if x+2<0 (i.e., x<−2){ |x + 2| = \begin{cases} x + 2 & \text{if } x + 2 \geq 0 \text{ (i.e., } x \geq -2)\\ -(x + 2) & \text{if } x + 2 < 0 \text{ (i.e., } x < -2) \end{cases} }

2. Splitting the Integral

Since the behavior of ∣x+2∣{ |x + 2| } changes at x=−2{ x = -2 }, we need to split the integral into two parts, one for the interval where x<−2{ x < -2 } and another for the interval where x≥−2{ x \geq -2 }. Our integral spans from −4{ -4 } to 0{ 0 }, and −2{ -2 } lies within this interval. Therefore, we split the integral as follows:

∫−40∣x+2∣dx=∫−4−2∣x+2∣dx+∫−20∣x+2∣dx{ \int_{-4}^{0} |x + 2| dx = \int_{-4}^{-2} |x + 2| dx + \int_{-2}^{0} |x + 2| dx }

3. Evaluating Each Integral

Now we replace ∣x+2∣{ |x + 2| } with its appropriate definition in each interval:

  • For −4≤x<−2{ -4 \leq x < -2 }: ∣x+2∣=−(x+2){ |x + 2| = -(x + 2) } ∫−4−2∣x+2∣dx=∫−4−2−(x+2)dx=−∫−4−2(x+2)dx{ \int_{-4}^{-2} |x + 2| dx = \int_{-4}^{-2} -(x + 2) dx = -\int_{-4}^{-2} (x + 2) dx } =−[x22+2x]−4−2=−[((−2)22+2(−2))−((−4)22+2(−4))]{ = -\left[\frac{x^2}{2} + 2x\right]_{-4}^{-2} = -\left[\left(\frac{(-2)^2}{2} + 2(-2)\right) - \left(\frac{(-4)^2}{2} + 2(-4)\right)\right] } =−[(2−4)−(8−8)]=−(−2)=2{ = -\left[(2 - 4) - (8 - 8)\right] = -(-2) = 2 }

  • For −2≤x≤0{ -2 \leq x \leq 0 }: ∣x+2∣=x+2{ |x + 2| = x + 2 } ∫−20∣x+2∣dx=∫−20(x+2)dx{ \int_{-2}^{0} |x + 2| dx = \int_{-2}^{0} (x + 2) dx } =[x22+2x]−20=[(022+2(0))−((−2)22+2(−2))]{ = \left[\frac{x^2}{2} + 2x\right]_{-2}^{0} = \left[\left(\frac{0^2}{2} + 2(0)\right) - \left(\frac{(-2)^2}{2} + 2(-2)\right)\right] } =[0−(2−4)]=−(−2)=2{ = \left[0 - (2 - 4)\right] = -(-2) = 2 }

4. Summing the Results

Finally, we add the results of the two integrals:

∫−40∣x+2∣dx=∫−4−2∣x+2∣dx+∫−20∣x+2∣dx=2+2=4{ \int_{-4}^{0} |x + 2| dx = \int_{-4}^{-2} |x + 2| dx + \int_{-2}^{0} |x + 2| dx = 2 + 2 = 4 }

Therefore, the value of the definite integral is:

∫−40∣x+2∣dx=4{ \int_{-4}^{0} |x + 2| dx = 4 }

Conclusion

In this article, we successfully solved two distinct integration problems. First, we tackled the integral of a rational function, ∫x3−1x3−xdx{ \int \frac{x^3 - 1}{x^3 - x} dx }, using partial fraction decomposition. This technique involves rewriting the rational function as a sum of simpler fractions, making it easier to integrate. We found the solution to be x+ln⁡∣xx+1∣+C{ x + \ln\left|\frac{x}{x + 1}\right| + C }. Second, we evaluated the definite integral ∫−40∣x+2∣dx{ \int_{-4}^{0} |x + 2| dx }, which involved an absolute value function. This required us to split the integral into intervals where the expression inside the absolute value had a consistent sign. We found the value of the definite integral to be 4. These examples illustrate important techniques in integral calculus, providing a solid foundation for tackling more complex problems.