In Exercises 21–32, express the integrand as a sum of partial ...

Question

In Exercises 21–32, express the integrand as a sum of partial fractions and evaluate the integrals.

∫ (8x² + 8x + 2) / (4x² + 1)² dx

Accepted Answer

Hello. In this video, we are going to be evaluating the following integral. We are given the integral of 10 x2 + 10 x + 2, and this will be divided by the quantity 5 x2 + 1 quantity squared. Now, in order to approach this problem, we want to go ahead and simplify the integrand by using partial fraction decomposition. If we rewrite the instagram on the side, we have, again, 10 x 2 + 10 x + 2 divided by the quantity 5 X2 + 1 quantity squared. In the denominator we have a repeated quadratic equation and so our partial fractions are going to be defined as a x plus b divided by 5 x 2 + 1 plus c x + D divided by the quantity 5 x 2 + 1 quantity squared. Then if we were to multiply by the least common denominator, we can further simplify this equation to be 10 x 2 + 10 x + 2 equal to the quantity. A X + B. Multiplied by 5 x 2 + 1. Plus the quantity c x + d Now, if we were to multiply and group up like terms together on the right hand side, we can further rewrite this equation to be 10 x2 + 10 x + 2. And this is going to equal to 5 A X cubed plus 5B X 2 plus the quantity A + C multiplied by X plus the constant terms B + D. Now we can create a system of linear equations by setting the coefficients of both sides equal to each other. So, if we were to set the cube terms equal to each other on both sides of the equation, the first coefficient equation we will get is 5A equal to 0, and this is going to give us the coefficient A is equal to 0. Setting the x 2 coefficients equal to each other will give us 5b is equal to 10, and this will give us the value b is equal to 2. For the third equation, setting the x terms equal to each other will give us a plus c is equal to 10. And since we know that the value of A is 0, that is going to give us C is equal to 10. And for the constant terms, B plus D is going to equal to 2, but since we know that B is equal to 2, we are going to get d is equal to 0. So now if we were to plug these back into the partial fractions, our partial fractions are going to simplify to be 2 divided by 5 X2 + 1. Plus 10 x divided by the quantity 5 x 2 + 1 quantity squared. So this is going to be the simplified form of our integrant, but now what we want to do is we want to go ahead and assign each of these fractions their own integral symbol. So now what we're going to do is we're going to evaluate each of the integrals independently. So let's go ahead and start with the first integral. Now again, the first integral we have is 2 divided by 5 X2 + 1 with respect to X. Now here, what we can go ahead and do is rewrite the denominator. 5 x 2 is equivalent to the quantity squared of 5 x raised to the power of 2. Plus one. And here we can simplify this integral by using a U substitution. We will allow you to equal to the square root of 5 x. And that will give us duU is equal to the square root of 5 dx. So by applying this substitution, we can rewrite the first integral by moving 2 to the front of the front of the integral, we will have 2 multiplied by the square root of 5. Multiplied by the integral of 1 divided by u2 + 1. Oh, and I made a small error here, apologies, but since the square root of 5 is a constant value, we would have to divide it to the left of the DU statement and get 1 divided by the square root of 5 DU equal to DX. So the coefficient in front of this integral should be 2 divided by the square root of 5. But now what we have in the integral is 1 divided by u2 + 1. By using our tangent inverse identity, we can rewrite this integral to be 2 divided by the square root of 5 multiplied by the tangent inverse of u. By rationalizing and resubstituting our value of U, we can rewrite the value of the first integral to be 2 multiplied by the square root of 5 divided by 5 multiplied by the tangent inverse of the square root of 5 x. This is going to be the result of the first integral. Now we want to go ahead and evaluate the 2nd integral. Now the second integral was the integral of 10 x. Divided by 5 X2 + 1 quantity 2. Here, if we use a U substitution, we will allow you to equal to 5 x2 + 1 and DU will equal to 10 xDX. So by applying this substitution, we can rewrite the integral to be the integral of 1 divided by U 2. And the integral of 1 divided by u squad we evaluate to be -1 divided by u. Then by back substituting our U substitution into terms of x we will have -1 divided by 5 x 2 + 1, which is going to be the result of the second integral. So now that we've evaluated both of the integrals, we want to combine the results together. So doing so is going to give us 2 square root of 5 divided by 5, multiplied by tangent inverse of square root of 5 multiplied by X minus 1 divided by 5 x 2 plus 1, and because we were originally working with an undefinite integral, we will add a generic constant C. This is going to be the value of the given integral. And with that being said, the overall solution to this problem. Well this will be the solution to the problem. So I hope this video helps you in understanding on how to approach this problem, and we will go ahead and see you all in the next video.

In Exercises 21–32, express the integrand as a sum of partial fractions and evaluate the integrals.∫ (8x² + 8x + 2) / (4x² + 1)² dx

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In Exercises 21–32, express the integrand as a sum of partial fractions and evaluate the integrals.
∫ (8x² + 8x + 2) / (4x² + 1)² dx