Evaluate the integrals in Exercises 9–28. It may be necessary ...

Question

Evaluate the integrals in Exercises 9–28. It may be necessary to use a substitution first.

∫ [(x + 1) / (x² (x − 1))] dx

Accepted Answer

Welcome back everyone. Evaluate the integral 2 x + 3 divided by x2 x + 1 dx. For this problem we're going to begin by using the partial fraction decomposition. So we have our expression to x + 3 divided by x2, which is multiplied by x + 1. So now let's split this fraction into a divided by first of all x plus b divided by x2, and finally we're going to add c divided by the final linear factor which is x + 1. Now, let's go ahead. And find the common denominator, we already know what it is. So for the first fraction, we want to multiply both sides by x and x plus 1, which gives us a x multiplied by x + 1. For the second fraction, we're missing x + 1, so we're adding plus B multiplied by x + 1. And for the final fraction, we want to multiply both sides by x2. So plus C X 2. And this is divided by the common denominator x 2. X + 1 Now what we know is that the numerator is equal to 2 x plus 3. So what we can do is distribute the terms. We get a x 2 plus a x plus bx plus B plus c x 2, and this is equal to 2 x + 3. So now we can introduce a system of equations, but before we do that, let's just factor out the terms. So In addition to what we already did, we can. Factor x2, which gives us a plus c multiplied by x2. Plus a plus b multiplied by x + b and this is equal to 2 x + 3. And from here, let's add our system of equations. So first of all, A + C. Must be equal to 0 because we don't have any x squared on the right hand side. A plus b is going to be the coefficient in front of x, which is 2. So a plus b equals 2, and finally b is the free coefficient. B equals 3. Now that we know that B equals 3, using the second equation, we can show that A is equal to 2 minus B, so that's 2 minus 3, and this is -1. And from the first equation C equalsa, which is of -1, and that's 1. So now we can rewrite our expression as a divided by x, that's -1 divided by x plus b divided by x2. So that's + 3 divided by x2 plus C divided by x + 1. So plus 1 divided by x + 1. And we're going to integrate these three functions. What we can do for now is simply Remove the work that we have done and start integrating. So the first integral is going to be the integral of negative. 1 divided by XD X. For the second integral again we can factor out the constant plus 3 and integrate 1 divided by x 2 d x. And then we're going to add the integral of 1 divided by x plus 1 dx. Let's integrate each function. Starting with The integral of 1 divided by x dx, this is an elementary integral and its value is l of the absolute value of x. Let's say constant of integration. Let's call it C1. For the next integral we are going to integrate 1 divided by x 2, which is x to the power of -2 dx, and using the power rule that's x to the power of -1 divided by -1, or basically x to the power of -1 plus a constant of integration c2, and we can also rewrite it as -1. Divided by x plus e constant of integration c2. And for the final one, we have 1 divided by x plus 1 dx. Which is equal to ln of the absolute value of x + 1. Plus a constant of integration c 3. And now that we have these values, we can simply substitute and we're going to get negative. Lm of the absolute value of x. + 3. Multiplied by. -1 divided by x, let's omit the constants for now. And plus Ellen. Of the absolute value of x + 1. And we can add a constant of integration C, which will be a combination of those constants, right? Now let's simplify using the properties of logarithms. We can write l and of the absolute value of x + 1 divided by x. That's because we have a difference of two natural logarithms minus 3 divided by x plus c, which is basically our final answer for this problem. Thank you for watching.

Evaluate the integrals in Exercises 9–28. It may be necessary to use a substitution first.∫ [(x + 1) / (x² (x − 1))] dx

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Evaluate the integrals in Exercises 9–28. It may be necessary to use a substitution first.
∫ [(x + 1) / (x² (x − 1))] dx