### Video Transcript

Use partial fractions to evaluate the indefinite integral of dπ₯ over π₯ times π₯ squared plus one squared.

Weβre told to use partial fractions to evaluate our integral. What that means is weβre going to write one over π₯ times π₯ squared plus one all squared, as the sum of two or more slightly simpler rational functions. Now, we need to be a little bit careful here. One of the factors in our denominator is a quadratic. Itβs π₯ squared plus one. And this itself is being squared. So because itβs being squared, because we have a repeated factor, we need both π₯ squared plus one and π₯ squared plus one squared as two of the denominators on our rational functions. But also because this factor is a quadratic, the numerator needs to be of this form, of π΅π₯ plus πΆ or π·π₯ plus πΈ.

Now here, π΄, π΅, πΆ, π·, and πΈ are constants that we need to work out. So how do we do that? Well, we try to make the expression on the right look like the expression on the left. And so we create a common denominator of π₯ times π₯ squared plus one all squared. To achieve this, we multiply the numerator and denominator of our first fraction by π₯ squared plus one all squared, of our second fraction by π₯ times π₯ squared plus one, and of our third fraction by π₯. Now, the denominators are equal and we can add the numerators. So the numerator becomes π΄ times π₯ squared plus one all squared plus π₯ times π΅π₯ plus πΆ times π₯ squared plus one plus π₯ times π·π₯ plus πΈ. And of course, this is equal to our original fraction.

And so because the denominators of both our fractions are equal, this tells us that the numerators themselves must also be equal. That is, one must be equal to π΄ times π₯ squared plus one squared plus π₯ times π΅π₯ plus πΆ times π₯ squared plus one plus π₯ times π·π₯ plus πΈ. Now, to find out the values of the constants, notice how the second two terms have a factor of π₯. And so if we let π₯ be equal to zero in this equation, those two terms are essentially going to disappear. In fact, we get one equals π΄ times zero squared plus one squared plus zero plus zero. And of course, thatβs zero. And actually, this simplifies really nicely. And we find that π΄ is equal to one.

Letβs now replace π΄ with one in our earlier equation. Next, we distribute our parentheses. And Iβm going to order the terms by their exponents. When we do, we get π₯ to the fourth power plus π΅π₯ to the fourth power plus πΆπ₯ cubed plus two π₯ squared plus π΅π₯ squared plus π·π₯ squared plus πΆπ₯ plus πΈπ₯ plus one. Weβre now going to equate coefficients of π₯ on the left- and right-hand side. Weβll begin by comparing coefficients of π₯ to the power of four. And on the left-hand side, there are no terms with that power, so we write zero. And on the right, we have π₯ to the fourth power plus π΅π₯ to the fourth power. So the coefficients there are one plus π΅.

If we subtract one from both sides, we find that π΅ is equal to negative one. Letβs repeat this process with π₯ cubed. On the left, we have zero and, on the right, we have πΆ. So actually, πΆ is equal to zero. Then for π₯ squared, we get zero equals two plus π΅ plus π·. But of course we saw that π΅ was equal to negative one. And if we simplify and solve for π·, we find that π· is equal to negative one, too. Finally, we compare coefficients of π₯ or π₯ to the power of one. We get zero equals πΆ plus πΈ. But we know that πΆ is zero and so πΈ is also zero. We now replace π΄, π΅, πΆ, π·, and πΈ in our earlier fractions with these values.

Now that weβve done so, we can clear some space and evaluate our integral. Weβre going to integrate one over π₯ minus π₯ over π₯ squared plus one minus π₯ over π₯ squared plus one all squared with respect to π₯. Well, the integral of one over π₯ is the natural log of the absolute value of π₯. But what about the other two integrals? Well, letβs integrate negative π₯ over π₯ squared plus one. Weβll do this by letting π’ be equal to π₯ squared plus one. This means dπ’ by dπ₯ is equal to two π₯. And we can, therefore, say even though dπ’ by dπ₯ isnβt a fraction, we do treat it a little like one. And a half dπ’ equals π₯ dπ₯.

By replacing π₯dπ₯ with a half dπ’ and π₯ squared plus one with π’, we can see that weβre looking to integrate negative one over two π’ with respect to π’. Well, thatβs negative one-half the natural log of the absolute value of π’. And of course, if we then replace π’ with π₯ squared plus one, we see that when we integrate this term, we get negative a half the natural log of the absolute value of π₯ squared plus one. Weβre actually going to perform the exact same substitution for our third term. This time, though, when we do the substitution, we get the integral of negative one over two π’ squared with respect to π’.

Letβs write this as negative one-half π’ to the power of negative two. When we integrate this, we get negative a half times π’ to the power of negative one divided by negative one. Thatβs a half π’ to the power of negative one or one over two π’. And so we replace π’ with π₯ squared plus one. And our third term integrates to one over two times π₯ squared plus one. Now, remember, this is an indefinite integral. So we add that constant of integration π. All thatβs left to do is to distribute the parentheses in our third term. And weβve evaluated our integral. Itβs the natural log of the absolute value of π₯ minus a half times the natural log of the absolute value of π₯ squared plus one plus one over two π₯ squared plus two plus the constant of integration π.