In Exercises 9–16, express the integrand as a sum of partial f...

Question

In Exercises 9–16, express the integrand as a sum of partial fractions and evaluate the integrals.

∫ (y + 4) / (y² + y) dy from 1/2 to 1

Accepted Answer

Welcome back everyone. Evaluate the integral from 1 to 2 of X + 5 DX divided by X2 + X. For this problem, first of all, let's consider the function that we're integrating that's x + 5 divided by x2 + x. We're going to use partial fractions here, so let's go ahead and factor out the denominator. That'd be x + 5 divided by x x + 1. So we can now rewrite our function as A divided by the first linear factor which is x plus b divided by the second linear factor which is x + 1. Now let's find a common denominator, which is going to be x multiplied by x + 1, and this means that we're multiplying a by x + 1. And adding b multiplied by x to get the numerator, and we know that our numerator is equal to x + 5. Now we can expand the left hand side a x plus a using the distributive property. Plus b x is equal to x + 5. And if we factor out x, we're going to get a plus b multiplied by x plus a equals 5. Now we can introduce a system of equations. A plus b is going to be equal to. One, so on the right hand side, let's make a correction. We know that our numerator is X + 5, right? So A plus B multiplied by X plus A is equal to x + 5. So a plus b must be equal to 1. That's the coefficient in front of x. And a itself is equal to the free coefficient which is 5. So if a is 5 from the first equation, we can show that b is equal to 1 minus a, which is 1 minus 5, and that's -4. Well done, so now we can show that the function that we are integrating can be represented as 5 divided by x. Plus B, which is negative force, we can just say -4 divided by x + 1. Great, and now we can split our integral into two parts. We're going to integrate from 1 to 2. The first function is 5 divided by x, and we can factor out 5. So we're going to get 1 divided by X D X. And we're going to subtract for again we can factor out the constant. Integral from 1 to 2. Of 1 divided by x plus 1 d x. Well then, in these are elementary functions. For the first part, we're going to get 5 LM. Of the absolute value of X, but since X is greater than 1, right? Essentially we don't have to include the absolute value in particular. X is greater than or equal to 1, right? We have the lower limit of 1 and the upper limit of 2. So x is between 1 and 2 inclusive, meaning we can omit the absolute value. We're going to have 5 ln of x minus for the second integral we get 4 ln of x plus 1. And now let's add to the limits of integration. That would be from 1 to 2, and let's apply the fundamental theorem of calculus, part. 2. There's going to be 5, a line of 2. -4 ln of 2 + 1, which is 3. Minus In parenthes. 5 ln of 1. -4 ln of 1 + 1 gives us 2, right? Now ln of 1 is 0, so we can cross out one term and then we can simplify using the properties of logarithms. So 5l of 2 can be written as ln of 2 to the power of 5, followed by -4 ln of 3, which is minus. Allen of 3 to the power of 4. And then if we distribute the negative sign, we're going to get negative of negative for alon of t, which is plus. For a line of 2 or a line of 2 to the power of 4. Now let's combine those. We're going to get Alan off. We can begin with 2 of 5, right? Multiplied by 2 to the power of 4 because we have plus 1 of 2 to the power of 4, and because we're subtracting elon of 3 to the power of 4, we're going to divide by 3 to the power of 4. And now this is equal to LN of 512 divided by 81. Which is our final answer for this problem. Thank you for watching.

In Exercises 9–16, express the integrand as a sum of partial fractions and evaluate the integrals.∫ (y + 4) / (y² + y) dy from 1/2 to 1

Watch next

In Exercises 9–16, express the integrand as a sum of partial fractions and evaluate the integrals.
∫ (y + 4) / (y² + y) dy from 1/2 to 1