Recall from algebra that a rational function f(x) is a quotient of two polynomial functions.

If the degree of the polynomial of the numerator is less than that of the denominator, then f(x) is a proper rational function.

The function is considered improper if the degree of the numerator is greater than that of the denominator.

In such a case, we can divide the numerator by denominator until we obtain a proper rational function.

Now, consider this indefinite integral of a rational function.

After making the u substitution, we are left with a simple integral.

The solution of this integral is straightforward.

Substituting for u gives us the general solution of the original integral.

Let's look at the integral of a different rational function. It's not possible to solve this integral using just a simple u substitution.

We must make use of a technique known as integration by partial fractions.

We must express the integrand of the given integral as a sum of simpler fractions.

These simple fractions are called partial fractions of the integrand.

We obtain these partial fractions by factoring the denominator of the original rational function.

We now have an integral that is easier for us to evaluate.

The integration of the partial fractions is straightforward. The decomposition of the original rational function was the key step. It is also the most difficult step.

This can be illustrated with a simple diagram. Here we have two irregular pieces that form a rectangle.

Given the two pieces, it is easy to put them together to form the rectangle.

However, given the rectangle, it is much more difficult determining the proper two pieces that make up the rectangle.

We will first look at a technique for determining partial fractions when the denominator of the rational function can be broken down into linear factors.

Consider this rational function. We wish to evaluate the indefinite integral of this function, but first we must break down the function into partial fractions.

We begin by factoring the denominator into two linear terms.

Next, we write the function as a sum of two partial fractions, each with a constant term in the numerator, and one of the linear factors in the denominator.

Now, we need to determine the values of the constants.

We multiply both sides of the expression by the product of the factors.

We then choose a value of x such that second term is zero.

Now, we can solve for the value of the constant in the first term.

We repeat the procedure by eliminating the first term, and then solving for the constant in the second term.

We can use the resulting partial fractions to evaluate the indefinite integral.

In the previous example, each factor in the denominator was different.

However, sometimes factors may be repeated. Consider this rational function.

Factoring the denominator, we see that there is one simple factor and one factor that is repeated twice.

Writing the function as a sum of partial fractions, we make one term with a constant in the numerator, and the simple factor in the denominator.

In the case of the repeated factor, we write it in terms of two partial fractions. One has a constant in the numerator, and a single factor in the denominator.

The other fraction has a constant in the numerator, and a squared factor in the denominator.

Our original function now is broken down into a total of three partial fractions.

In general, if a factor is repeated n times, it is broken down into n partial fractions.

We solve for the constants as before. First, we multiply both sides of the expression by the product of the original factors.

Next, we eliminate the second and third terms, and then solve for the first constant.

We see that we can determine the third constant by eliminating the first and second terms.

We can express our original function by the partial fractions we determined.

Now, we are able to evaluate the integral of the function.

Let's look at another example. Suppose a chemical reaction proceeds by the given mechanism. One molecule of C reacts with one molecule of D to form one molecule of X.

The rate at which X is formed is modeled by the given differential equation, and is illustrated by the graph.

We wish to determine the time t as a function of X by solving the differential equation. We do this by rearranging the differentials and taking the integral of both sides.

The integral on the left-hand side of the expression is easily solved.

We now have t in terms of X and the constant c.

We know that when t is zero, X is also zero.

We now have an expression for t in terms of X.

The graph illustrates how C, D and X change with respect to time.

We can solve it using a u substitution. What should we let u be equal to? Click the "Submit" button after selecting your answer.

What is the general solution of this integral? Click the "Submit" button after selecting your answer.

What is the general solution of this integral? Click the "Submit" button after selecting your answer.

What is the value of the second constant? Click the "Submit" button after selecting your answer.

What is the general solution of this indefinite integral? Click the "Submit" button after selecting your answer.

What is the solution of the integral on the right-hand side? Click the "Submit" button after selecting your answer.

What is the value of c? Click the "Submit" button after selecting your answer.

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