Step-by-Step Guide To Solving Integrals

Hey guys! Today, we're diving into the fascinating world of integrals. We'll break down a couple of examples step-by-step, making sure you grasp the underlying concepts. Let's get started!

1. Integrating a Rational Function Using Partial Fractions

First up, we have an integral involving a rational function:

$$\int \frac{5}{x^2+2 x-8} d x$$

The initial problem presents us with integrating a rational function, and our goal is to simplify this complex fraction into manageable parts. The given form suggests using partial fraction decomposition. This technique is super useful when dealing with rational functions, which are essentially fractions where both the numerator and denominator are polynomials.

To kick things off, we need to factor the denominator. Factoring the quadratic expression x² + 2x - 8 is crucial. We're looking for two numbers that multiply to -8 and add up to 2. Those numbers are 4 and -2. So, we can rewrite the denominator as (x + 4)(x - 2). This factorization allows us to express the original fraction as a sum of simpler fractions, a process known as partial fraction decomposition. This method simplifies the integration process significantly by breaking down a complex rational function into simpler terms that are easier to integrate.

Now, we can express the integrand as:

$$\frac{5}{x^2+2 x-8} = \frac{5}{(x+4)(x-2)} = \frac{A}{x+4} + \frac{B}{x-2}$$

Here, A and B are constants that we need to determine. This is the heart of partial fraction decomposition – breaking down a complex fraction into simpler fractions with unknown constants in the numerators. The goal is to find the values of these constants, which will allow us to rewrite the original integral as a sum of simpler integrals.

To find A and B, we multiply both sides of the equation by the common denominator (x + 4)(x - 2). This clears the fractions and gives us a simpler equation to work with:

$$5 = A(x-2) + B(x+4)$$

This equation holds true for all values of x. We can use strategic values of x to eliminate one variable at a time and solve for the unknowns. For example, if we set x = 2, the term with A becomes zero, allowing us to solve for B. Similarly, setting x = -4 eliminates the term with B, enabling us to solve for A.

Let's substitute x = 2:

$$5 = A(2-2) + B(2+4)$$

$$5 = 6B$$

$$B = \frac{5}{6}$$

Now, let's substitute x = -4:

$$5 = A(-4-2) + B(-4+4)$$

$$5 = -6A$$

$$A = -\frac{5}{6}$$

Fantastic! We've found the values of A and B. Now we can rewrite the original integral as:

$$\int \frac{5}{x^2+2 x-8} d x = \int \left( \frac{-5/6}{x+4} + \frac{5/6}{x-2} \right) d x$$

See how much simpler this looks? This step is crucial because it transforms the original complex integral into a sum of simpler integrals that we can easily solve using basic integration rules. The constants A and B act as coefficients for these simpler integrals, making the integration process straightforward.

Now we can split the integral and pull out the constants:

$$= -\frac{5}{6} \int \frac{1}{x+4} d x + \frac{5}{6} \int \frac{1}{x-2} d x$$

These integrals are standard forms. The integral of 1/(x + a) is ln|x + a| + C, where C is the constant of integration. Applying this rule to both integrals, we get:

$$= -\frac{5}{6} \ln|x+4| + \frac{5}{6} \ln|x-2| + C$$

We can combine the logarithms using the logarithm property ln(a) - ln(b) = ln(a/b):

$$= \frac{5}{6} \ln\left|\frac{x-2}{x+4}\right| + C$$

And there you have it! We've successfully integrated the first function using partial fraction decomposition. Remember, the key is to break down the complex rational function into simpler fractions, making the integration process much easier. This technique is widely applicable in calculus and is a valuable tool for solving a variety of integration problems.

2. Integrating a Rational Function by Simplifying

Next, let's tackle this integral:

$$\int \frac{x^2-x+1}{x^2} d x$$

In this case, we're faced with integrating another rational function, but this time, the approach is slightly different. Instead of using partial fractions directly, we can simplify the integrand first. Sometimes, a little algebraic manipulation can make a big difference in how easy an integral is to solve. Always be on the lookout for opportunities to simplify before diving into more complex techniques.

We can divide each term in the numerator by x²:

$$\int \frac{x^2-x+1}{x^2} d x = \int \left( \frac{x^2}{x^2} - \frac{x}{x^2} + \frac{1}{x^2} \right) d x$$

Simplifying each fraction, we get:

$$= \int \left( 1 - \frac{1}{x} + \frac{1}{x^2} \right) d x$$

This looks much more manageable, doesn't it? By breaking the fraction down into individual terms, we've transformed the integral into a sum of simpler integrals that we can easily handle using basic integration rules. This is a common strategy in calculus – simplifying the integrand to make the integration process more straightforward.

Now, we can split the integral into three separate integrals:

$$= \int 1 d x - \int \frac{1}{x} d x + \int \frac{1}{x^2} d x$$

Each of these integrals is a standard form that we should be familiar with. The integral of 1 is simply x, the integral of 1/x is ln|x|, and the integral of 1/x² is -1/x. Remember, practice makes perfect when it comes to recognizing these standard forms. The more you encounter them, the quicker you'll be able to identify and integrate them.

Let's integrate each term:

$$= x - \ln|x| + \int x^{-2} d x$$

For the last term, we use the power rule for integration, which states that the integral of xⁿ is (x^(n+1))/(n+1) + C, where C is the constant of integration. Applying this rule, we get:

$$= x - \ln|x| + \frac{x^{-1}}{-1} + C$$

Simplifying, we have:

$$= x - \ln|x| - \frac{1}{x} + C$$

And that's it! We've successfully integrated the second function by simplifying the integrand first. This approach highlights the importance of looking for opportunities to simplify before applying more complex integration techniques. Sometimes, a little algebraic manipulation can save you a lot of time and effort.

Conclusion

So, there you have it! We've worked through two different integral problems, showcasing different techniques. Remember, the key to mastering integration is practice, practice, practice! Keep exploring, keep learning, and you'll become an integration pro in no time. Keep these strategies in your toolbox, and you'll be well-equipped to tackle a wide range of integration problems. Happy integrating!