# Video: Integrating Using Partial Fractions

Find πΉ so that πΉβ²(π₯) = π₯/(π₯ β π)Β² and πΉ(0) = 2.

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### Video Transcript

Find the function πΉ so that πΉ prime of π₯ is equal to π₯ divided by π₯ minus π squared and πΉ of zero is equal to two.

The question gives us the derivative function of πΉ and an initial value for our function πΉ. We can find our function πΉ as itβs the antiderivative of our function πΉ prime of π₯. And weβll use our initial value that πΉ evaluated at zero is equal to two.

However, trying to do this directly leads to a problem. We canβt directly integrate π₯ divided by π₯ minus π all squared. So weβll have to manipulate this into an easier-to-integrate function. Weβll do this by using partial fractions.

Since we have a repeated root in our denominator, partial fractions gives us that π₯ divided by π₯ minus π squared is equal to some constant π΄ divided by π₯ minus π. Plus some constant π΅ divided by π₯ minus π squared. To help us find the values of π΄ and π΅, weβll multiply both sides of our equation by π₯ minus π squared. This gives us that π₯ is equal to π΄ multiplied by π₯ minus π plus π΅. And since this is true for all values of π₯, weβll use an equivalent sign.

We can eliminate the variable π΄ by using the substitution π₯ is equal to π. This gives us π is equal to π΄ multiplied by π minus π plus π΅. And since π minus π is just equal to zero, we have that π΅ is equal to π.

Now weβll use the fact that π is equal to π΅ and the substitution π₯ is equal to zero. To see that zero is equal to π΄ multiplied by zero minus π plus π. This gives us that zero is equal to negative π΄π plus π. We add π΄π to both sides to get that π΄π is equal to π.

Finally, we divide both sides of our equation by π to get that π΄ is equal to one. So weβve shown by partial fractions that our derivative function πΉ prime of π₯, which is equal to π₯ divided by π₯ minus π squared. Is equal to one divided by π₯ minus π plus π divided by π₯ minus π squared. So weβre now ready to start calculating the antiderivative of our derivative function πΉ prime of π₯.

Weβll have that πΉ of π₯ is equal to the integral of πΉ prime of π₯ with respect to π₯ up to a constant of integration. Which is equal to the integral of π₯ divided by π₯ minus π squared with respect to π₯. By partial fractions, we have that this is equal to the integral of one divided by π₯ minus π plus π divided by π₯ minus π squared with respect to π₯.

The easiest way to continue with this integral is to notice that both of our denominators are in terms of π₯ minus π. So weβll use the substitution π’ is equal to π₯ minus π. Differentiating both sides of this equation with respect to π₯ gives us that dπ’ by dπ₯ is equal to one. And we get the equivalent statement dπ’ is equal to dπ₯.

So by using the substitution π’ is equal to π₯ minus π, we now have the integral of one divided by π’ plus π divided by π’ squared with respect to π’. To evaluate the first term in our integral, we recall that the integral of one over π₯ with respect to π₯ is equal to the natural logarithm of the absolute value of π₯ plus the constant of integration πΆ.

And to evaluate the second term in our integral, we recall that, for constants π΄ and π, where π is not equal to negative one. The integral of π΄ multiplied by π₯ to the πth power with respect to π₯ is equal to π΄ multiplied by π₯ to the power of π. Plus one divided by π plus one plus the constant of integration πΆ.

We also notice that our second term is equal to π multiplied by π’ to the power of negative two. So integrating one over π’ with respect to π’ gives us the natural logarithm of the absolute value of π’. And integrating π divided by π’ squared with respect to π’ gives us negative π divided by π’. And instead of adding two constants of integration, weβll just add one constant of integration we will call πΆ.

Now we substitute back in that π’ was equal to π₯ minus π. To get the natural logarithm of the absolute value of π₯ minus π minus π divided by π₯ minus π plus our constant of integration πΆ. And we recall this is our antiderivative function πΉ of π₯ up to our constant of integration πΆ.

Well, the question tells us that πΉ evaluated at zero is equal to two. So substituting π₯ is equal to zero gives us that πΉ evaluated at zero is equal to the natural logarithm of the absolute value of zero. Minus π minus π divided by zero minus π plus the constant of integration πΆ.

The question tells us that πΉ evaluated at zero is equal to two. The absolute value of zero minus π is equal to π. So the natural logarithm of π is just equal to one. Then we subtract π divided by zero minus π. So weβre subtracting negative one, which is adding one. Then we add our constant of integration πΆ. Then we subtract two from both sides of our equation to see that πΆ is equal to zero.

Therefore, since πΆ is equal to zero and negative π divided by π₯ minus π is equal to π divided by π minus π₯. We have that πΉ of π₯ is equal to the natural logarithm of the absolute value of π₯ minus π plus π divided by π minus π₯.