Question

$$\text {Evaluate the following integrals.}$$ $$\int \frac{x-5}{x^{2}(x+1)} d x$$

Short answer

Question: Evaluate the integral of the given function: $$\int \frac{x-5}{x^{2}(x+1)} dx$$ Answer: $$\int \frac{x-5}{x^{2}(x+1)} dx = -\frac{1}{2}\ln|x| + \frac{5}{x} + 6\ln|x+1| + C$$

Step-by-step solution

Apply partial fraction decomposition

To evaluate the integral, first, we need to apply partial fraction decomposition to the integrand. We can write the given function as follows: $$\frac{x-5}{x^{2}(x+1)} = \frac{A}{x} + \frac{B}{x^2} + \frac{C}{x+1}$$ where A, B, and C are constants to be determined. Now, multiplying both sides by \(x^2(x+1)\) to eliminate denominators: $$(x-5) = A x^2 (x+1) + B (x+1) + C x^3$$

Solve for A, B, and C

Now we'll find A, B, and C by solving the above equation. To find A, set x = 0: $$-5 = A(0)(1) + B(1) + C(0^3) \Rightarrow B = -5$$ To find C, set x = -1, and plug in B: $$(-1 - 5) = A((-1)^2)((-1) + 1) - 5 ((-1) + 1) + C((-1)^3) \Rightarrow C = 6$$ Now let's find A: $$x-5 = A x^3 (x+1) - 5x (x+1) + 6x^3$$ Plugging in the found values of B and C, and re-arranging this equation, we get: $$5x^3 - 5x^2 + A x^3 (x+1) = x-5$$ Setting x = 1, we can simplify this equation to find A: $$5 - 5 + A (1) (2) = 1 - 5 \Rightarrow A=- \frac{1}{2}$$ Now we have: $$\frac{x-5}{x^{2}(x+1)} = - \frac{1}{2x} - \frac{5}{x^2} + \frac{6}{x+1}$$

Integrate each fraction

Now we need to integrate each fraction separately: $$\int \frac{x-5}{x^{2}(x+1)} dx = \int \left(-\frac{1}{2x} - \frac{5}{x^2} + \frac{6}{x+1} \right) dx$$ $$= - \frac{1}{2} \int \frac{1}{x} dx - 5 \int \frac{1}{x^2} dx + 6 \int \frac{1}{x+1} dx$$ Now let's integrate each term: $$= -\frac{1}{2} \ln|x| - 5 \left(-\frac{1}{x}\right) + 6 \ln|x+1| + C$$

Combine terms and write the final answer

Now we combine the integrated terms and write our final answer: $$\int \frac{x-5}{x^{2}(x+1)} dx = -\frac{1}{2}\ln|x| + \frac{5}{x} + 6\ln|x+1| + C$$ Here, C is the constant of integration.

Key concepts

Calculus

Calculus is a branch of mathematics focused on limits, functions, derivatives, integrals, and infinite series. In the realm of calculus, integration is a fundamental concept, which essentially examines the accumulation of quantities and the areas under and between curves.

One way to think about integration is that it 'reverses' differentiation. Where differentiation gives us the slope of a curve at any point, integration gives the total value accumulated by a function, such as the area under the curve, from a starting point to an ending point. This exercise involves finding the area under a given rational function, one of the quintessential applications of calculus.

Integral Evaluation

Integral evaluation is the process of determining the value of an integral, which often represents the area under a curve. In the context of our exercise, we are required to evaluate an indefinite integral, which entails finding the antiderivative of a function. An antiderivative of a function is another function whose derivative is the original function.

To solve an integral, particularly those that are not immediately straightforward, various techniques may be employed, including substitution, integration by parts, and partial fraction decomposition. Our exercise utilized partial fraction decomposition to break down a complex rational expression into simpler fractions that can be integrated separately and easily.

Integration Techniques

There are multiple techniques used to integrate complex functions, each one suitable for different scenarios. One such technique, used in our exercise, is partial fraction decomposition. This method is particularly useful when dealing with rational functions, where the numerator and the denominator are polynomials and the degree of the numerator is less than the degree of the denominator.

The idea is to break down the complex fraction into simpler parts that can be integrated individually. Once the constants for each term in the partial fraction decomposition are determined, the integral of the complex function becomes the sum of the integrals of these simpler fractions, making the evaluation of the integral a more approachable task.