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.

∫ (3t² + t + 4) / (t³ + t) dt from 1 to √3

Accepted Answer

Hello. In this video, we are going to be evaluating the following integral. We have the definite integral from 1 to 2 of 4 x 2 + 2 x + 3 divided by x cubed plus x. Now here we can simplify the integrand by using partial fraction decomposition. Now, if we were to rewrite, or if we were to factor in X out of the denominator, we can rewrite the integrand to be 4 X2 + 2 X + 3, divided by X multiplied by X2 + 1. Here we have a non-repeating linear term and a non-repeating quadratic term, so the partial fractions are going to be defined as a divided by x plus B X + C divided by x2 + 1. Now, if we were to multiply both sides of this equation by the least common denominator, we can rewrite this equation to be 4 X2 + 2 X + 3 equal to a multiplied by X2 + 1 plus the quantity BX + C multiplied by X. Now, what we're going to do in this case is we're going to expand the right hand side and reorder the terms. Doing so will allow us to rewrite the equation as 4 X2 + 2 X + 3. Equal to the quantity a + b multiplied by x2 + c x plus the constant value a. Now, what we want to do is we want to create a system of equations to solve for the missing coefficients A, B, and C. By setting the coefficients of x2 on both sides of the equation equal to each other, our first equation is going to be a + b is equal to 4. By repeating this process for the coefficients of the x terms, our second equation will be c is equal to 2, which is one of the coefficients we are looking for. And repeating this for the constant values is going to give us A is equal to 3, which is another constant value we are looking for. And since we know the value of A, plugging in the value of A into the first equation will leave us with 4 + 3 + B. Equal to 4 and solving for b is going to give us b is equal to 1. Now, if we were to take these coefficients of A, B, and C and plug them back into the partial fractions, the partial fractions can be redefined as 3 divided by X plus X + 2 divided by X2 + 1. We can further break apart the second fraction into a sum of fractions, and we can rewrite the partial fractions as 3 divided by x plus x divided by x2 + 1 + 2 divided by x2 + 1. Now, since the original integral had the definite boundaries from 1 to 2, each of our partial fractions are also going to have a definite boundary from 1 to 2. So what we need to do now is we need to go through each of the partial fractions and integrate them independently. So let's go ahead and start with the first partial fraction. Now, moving the coefficient of 3 to the front of the integral is going to leave us with 3 multiplied by the integral from 1 to 2 of 1 divided by XDX. By using our natural logarithm property, the antiderivative is going to be 3 multiplied by the natural logarithm of x, and this is going to be evaluated from 1 to 2. Evaluating this at the boundary is going to leave us with 3 multiplied by the natural logarithm of 2 minus 3 multiplied by the natural logarithm of 1. But the natural logarithm of 1 is going to simplify to be 0, so we are left with 3 multiplied by the natural logarithm of 2. Now by using the natural logarithm properties, we can move the 3 inside the natural logarithm as an exponent. This will give us the natural logarithm of 2 cubed which is then going to give us the natural logarithm of 8. So this is going to be the value of the first integral. Now let's go ahead and evaluate the 2nd integral. Now, the second integral is the integral from 1 to 2 of X divided by X2 + 1 DX. We can evaluate this integral by using a U substitution. We will allow you to equal to x2 + 1. Then DU will equal to 2 XDX, which can further be rewritten as 1/2 DU is equal to XDX. Now, this is going to allow us to rewrite the integral to be the integral of 1 divided by UDU. And since we performed a U substitution over a definite integral, we have to plug in our boundaries into the U substitution. Plugging in 1 into X2 + 1 is going to give us a lower boundary of 2. And plugging in 2 into x2 + 1 is going to give us an upper boundary of 5. Similarly, this is going to integrate to be the natural logarithm of you evaluated from 2 to 5. This will then further simplify. Oh, sorry, I forgot one thing. We have a coefficient of 1/22, so there should be a 1/2 in front of the substitution, which means that we are missing a 1/2 in front of our natural logarithm. But evaluating this at the boundary is going to leave us with 1/2 multiplied by the natural logarithm of 5 minus 1/2 multiplied by the natural logarithm of 2. Now by using our natural logarithm properties, this is going to further simplify to be 1/2 multiplied by the natural logarithm of 5 divided by 2. So this is going to be the value of the 2nd integral. Now we just need to evaluate the final integral. For the third integral, we have the integral from 2, from 1 to 2. Of 2 divided by x2 + 1 Now, if we were to move this coefficient of 2 in front of the integral, we have 2 multiplied by the integral from 1 to 2 of 1 divided by X2 + 1. By the tangent inverse identity, the antiderivative of this is going to be 2 multiplied by the tangent inverse of x divided by 2, and this will be evaluated from 1 to 2. This will further simplify to be 2 multiplied by tangent inverse of 1 or sorry tangent inverse of 2. Minus the tangent inverse of 1. And this will further simplify to be 2 multiplied by the tangent inverse of 2 minus pi divided by 4. And once we distribute this 2 into the quantity, we will have 2 multiplied by the tangent inverse of 2. Minus pi divided by 2. And this is going to be the value of the final integral. Now if we were to take the results of the three integrals and combine them back together, the overall solution is going to be the natural logarithm of 8 plus 1/2 multiplied by the natural logarithm of 5 divided by 2. Plus 2 multiplied by the tangent inverse of 2 minus pi divided by 2 and this is going to be the solution to the definite integral with the overall solution being option a. 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.∫ (3t² + t + 4) / (t³ + t) dt from 1 to √3

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In Exercises 21–32, express the integrand as a sum of partial fractions and evaluate the integrals.
∫ (3t² + t + 4) / (t³ + t) dt from 1 to √3