# Question Video: Integration by Partial Fractions

Find the Partial Fraction Decomposition – Super Easy Way
Find the Partial Fraction Decomposition – Super Easy Way

### Video Transcript

Use partial fractions to evaluate the integral, the indefinite integral of two 𝑥 over 𝑥 plus seven squared with respect to 𝑥.

We’re actually told to solve this problem by using partial fractions. Now remember, when we use partial fractions, we write our rational function as the sum of simpler rational expressions. Now in this case, we need to be a little bit careful because we have something that we call a repeated factor.

So when we write two 𝑥 over 𝑥 plus seven all squared using partial fractions, we write it as 𝐴 over 𝑥 plus seven plus 𝐵 over 𝑥 plus seven squared. If the denominator of our fraction had have been 𝑥 plus seven cubed. For example, we would’ve had 𝐴 over 𝑥 plus seven plus 𝐵 over 𝑥 plus seven squared plus 𝐶 over 𝑥 plus seven cubed.

Now our aim is to make sure that the expression on the right looks like that on the left. And so we remember that, to add two fractions, we create a common denominator. Here, to achieve that, we’re going to multiply 𝐴 over 𝑥 plus seven by the numerator and the denominator of that fraction by 𝑥 plus seven. When we do, the right-hand side becomes 𝐴 times 𝑥 plus seven over 𝑥 plus seven all squared plus 𝐵 over 𝑥 plus seven all squared.

And now that the denominators are equal, we add the numerators. And so we find that two 𝑥 over 𝑥 plus seven all squared can be written as 𝐴 times 𝑥 plus seven plus 𝐵 over 𝑥 plus seven all squared. Now notice that the denominators in our fractions are equal. And that means, for the fractions themselves to be equal, their numerators must also be equal. That is, two 𝑥 must be equal to 𝐴 times 𝑥 plus seven plus 𝐵. So how do we work out the values of the constants 𝐴 and 𝐵?

We have two methods. We can substitute in the roots of 𝑥 plus seven squared. Alternatively, we can distribute our parentheses and equate coefficients. Now we’re going to use the latter method here.

When we distribute our parentheses, on the right-hand side, we get 𝐴𝑥 plus seven 𝐴 plus 𝐵. We’ll begin by equating coefficients of 𝑥 or 𝑥 to the power of one. On the left-hand side, the coefficient of 𝑥 is two. And on the right-hand side, the coefficient of 𝑥 is 𝐴. So we find that two is equal to 𝐴.

Next, we equate coefficients of 𝑥 to the power of zero. Well, really, this is equating constants. On the left-hand side, the coefficient of 𝑥 to the power of zero or the constant is actually zero. And on the right-hand side, it’s seven 𝐴 plus 𝐵. But of course, we saw that 𝐴 is equal to two. So we can write this as zero equals seven times two plus 𝐵 or zero equals 14 plus 𝐵.

By subtracting 14 from both sides of this equation, we find 𝐵 to be equal to negative 14. And so by replacing 𝐴 with two and 𝐵 with negative 14, we need to evaluate the integral of two over 𝑥 plus seven minus 14 over 𝑥 plus seven squared with respect to 𝑥. Well, when we integrate one over 𝑥 plus some constant, we get the natural log of the absolute value of 𝑥 plus that constant. So the integral of one over 𝑥 plus seven is the natural log of the absolute value of 𝑥 plus seven.

This in turn means the integral of two over 𝑥 plus seven is two times the natural log of the absolute value of 𝑥 plus seven. But how do we integrate negative 14 over 𝑥 plus seven all squared? Well, we begin by taking out the constant factor of negative 14. This isn’t entirely necessary, but it does make the next step a little easier.

Next, we’re going to use integration by substitution. And the substitution we choose is 𝑢 equals 𝑥 plus seven. Differentiating 𝑢 with respect to 𝑥, d𝑢 by d𝑥, and we get equal to one. And whilst d𝑢 by d𝑥 isn’t a fraction, we treat it a little like one. And we can say that d𝑢 must be equal to d𝑥. And this means the integral of one over 𝑥 plus seven all squared d𝑥 is the same as the integral of one over 𝑢 squared d𝑢.

Now if we write one over 𝑢 squared is 𝑢 to the power of negative two, we find that the integral of 𝑢 to the power of negative two is 𝑢 to the power of negative one divided by negative one. Or negative 𝑢 to the power of negative one. Now we are dealing with an indefinite integral. So we’re going to have a constant of integration. We’re actually going to add that at the end.

Now if we think of negative 𝑢 to the power of negative one as negative one over 𝑢. And we then replace 𝑢 with 𝑥 plus seven. We find that the integral of one over 𝑥 plus seven squared with respect to 𝑥 is negative one over 𝑥 plus seven. And so we replaced that integral with negative one over 𝑥 plus seven in our expression for our earlier integral.

We distribute the parentheses. Negative 14 times negative one over 𝑥 plus seven is 14 over 𝑥 plus seven. And then we simply factor by two. And when we do, we see that the integral of two 𝑥 over 𝑥 plus seven squared with respect to 𝑥 is two times the natural log of the absolute value of 𝑥 plus seven plus seven over 𝑥 plus seven plus that constant of integration 𝑘.

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