Video: Integrating Using Partial Fractions

Find 𝐹 so that 𝐹′(π‘₯) = π‘₯/(π‘₯ βˆ’ 𝑒)Β² and 𝐹(0) = 2.

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Video Transcript

Find the function 𝐹 so that 𝐹 prime of π‘₯ is equal to π‘₯ divided by π‘₯ minus 𝑒 squared and 𝐹 of zero is equal to two.

The question gives us the derivative function of 𝐹 and an initial value for our function 𝐹. We can find our function 𝐹 as it’s the antiderivative of our function 𝐹 prime of π‘₯. And we’ll use our initial value that 𝐹 evaluated at zero is equal to two.

However, trying to do this directly leads to a problem. We can’t directly integrate π‘₯ divided by π‘₯ minus 𝑒 all squared. So we’ll have to manipulate this into an easier-to-integrate function. We’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 𝑒 squared is equal to some constant 𝐴 divided by π‘₯ minus 𝑒. Plus some constant 𝐡 divided by π‘₯ minus 𝑒 squared. To help us find the values of 𝐴 and 𝐡, we’ll multiply both sides of our equation by π‘₯ minus 𝑒 squared. This gives us that π‘₯ is equal to 𝐴 multiplied by π‘₯ minus 𝑒 plus 𝐡. And since this is true for all values of π‘₯, we’ll use an equivalent sign.

We can eliminate the variable 𝐴 by using the substitution π‘₯ is equal to 𝑒. This gives us 𝑒 is equal to 𝐴 multiplied by 𝑒 minus 𝑒 plus 𝐡. And since 𝑒 minus 𝑒 is just equal to zero, we have that 𝐡 is equal to 𝑒.

Now we’ll use the fact that 𝑒 is equal to 𝐡 and the substitution π‘₯ is equal to zero. To see that zero is equal to 𝐴 multiplied by zero minus 𝑒 plus 𝑒. This gives us that zero is equal to negative 𝐴𝑒 plus 𝑒. We add 𝐴𝑒 to both sides to get that 𝐴𝑒 is equal to 𝑒.

Finally, we divide both sides of our equation by 𝑒 to get that 𝐴 is equal to one. So we’ve shown by partial fractions that our derivative function 𝐹 prime of π‘₯, which is equal to π‘₯ divided by π‘₯ minus 𝑒 squared. Is equal to one divided by π‘₯ minus 𝑒 plus 𝑒 divided by π‘₯ minus 𝑒 squared. So we’re now ready to start calculating the antiderivative of our derivative function 𝐹 prime of π‘₯.

We’ll have that 𝐹 of π‘₯ is equal to the integral of 𝐹 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 𝑒 squared with respect to π‘₯. By partial fractions, we have that this is equal to the integral of one divided by π‘₯ minus 𝑒 plus 𝑒 divided by π‘₯ minus 𝑒 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 𝑒. So we’ll use the substitution 𝑒 is equal to π‘₯ minus 𝑒. Differentiating both sides of this equation with respect to π‘₯ gives us that d𝑒 by dπ‘₯ is equal to one. And we get the equivalent statement d𝑒 is equal to dπ‘₯.

So by using the substitution 𝑒 is equal to π‘₯ minus 𝑒, we now have the integral of one divided by 𝑒 plus 𝑒 divided by 𝑒 squared with respect to 𝑒. 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 𝐢.

And to evaluate the second term in our integral, we recall that, for constants 𝐴 and 𝑛, where 𝑛 is not equal to negative one. The integral of 𝐴 multiplied by π‘₯ to the 𝑛th power with respect to π‘₯ is equal to 𝐴 multiplied by π‘₯ to the power of 𝑛. Plus one divided by 𝑛 plus one plus the constant of integration 𝐢.

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

Now we substitute back in that 𝑒 was equal to π‘₯ minus 𝑒. To get the natural logarithm of the absolute value of π‘₯ minus 𝑒 minus 𝑒 divided by π‘₯ minus 𝑒 plus our constant of integration 𝐢. And we recall this is our antiderivative function 𝐹 of π‘₯ up to our constant of integration 𝐢.

Well, the question tells us that 𝐹 evaluated at zero is equal to two. So substituting π‘₯ is equal to zero gives us that 𝐹 evaluated at zero is equal to the natural logarithm of the absolute value of zero. Minus 𝑒 minus 𝑒 divided by zero minus 𝑒 plus the constant of integration 𝐢.

The question tells us that 𝐹 evaluated at zero is equal to two. The absolute value of zero minus 𝑒 is equal to 𝑒. So the natural logarithm of 𝑒 is just equal to one. Then we subtract 𝑒 divided by zero minus 𝑒. So we’re subtracting negative one, which is adding one. Then we add our constant of integration 𝐢. Then we subtract two from both sides of our equation to see that 𝐢 is equal to zero.

Therefore, since 𝐢 is equal to zero and negative 𝑒 divided by π‘₯ minus 𝑒 is equal to 𝑒 divided by 𝑒 minus π‘₯. We have that 𝐹 of π‘₯ is equal to the natural logarithm of the absolute value of π‘₯ minus 𝑒 plus 𝑒 divided by 𝑒 minus π‘₯.

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