Evaluate the integrals in Exercises 69–134. The integrals are ...

Question

Evaluate the integrals in Exercises 69–134. The integrals are listed in random order so you need to decide which integration technique to use.

∫ (x + 1) / (x⁴ − x³) dx

Accepted Answer

Hello there. Today we are going to solve the following practice problem together. So first off, let us read the problem and highlight all the key pieces of information that we need to use in order to solve this problem. Evaluate the integral, the integral of x + 4 divided by x 4 minus x 3 dx. OK. So it appears for this particular prompt, we're asked to ultimately evaluate the integral that is provided to us. So now that we know that we're ultimately trying to evaluate this integral, we need to recall a note based on our prior knowledge that in order to evaluate the integral of x + 4 divided by x 4 minus x 3. DX, we will need to recall and use the method of partial fraction decomposition. So before we could implement partial fraction decomposition, our first step that we need to take is we need to simplify our current integral that is given to us by factoring out the denominator. So first we need to factor out the greatest common factor from the denominator, which we could take x 4 minus x 3, which will be equal to x 3 multiplied by parentheses of x minus 1. So that will mean our integrand will become X + 4 divided by X 3 multiplied by parentheses of X minus 1. Our second step now is we need to set up our partial fractions, and since we have a repeated linear factor of x 3. And a simple linear factor, which is parentheses x minus 1, we could write our decomposition as x + 4 divided by x to the power of 3 multiplied by parentheses of x minus 1. Which will be equal to a divided by x plus B divided by x 2 plus C divided by x 3 plus D divided by x minus 1. So, what we need to do now, and let's make a note off to the side, is we now need to multiply both sides of our expression by x 3 multiplied by parentheses of x minus 1, and when we do that, we will find that we could write x + 4 is equal to. A multiplied by x 2 multiplied by parentheses of x minus 1 + b multiplied by x multiplied by parentheses of x minus 1 plus c multiplied by parentheses of x minus 1 plus d multiplied by x to the power of 3. And we can simplify this expression to write x + 4 is equal to parentheses of a plus d multiplied by x to the power of 3. Plus parentheses B minus a multiplied by x 2. Plus parentheses C minus B multiplied by X minus C. OK, so our third step now is we need to solve for our constants, and in order to solve for D, we need to let X equal 1, so we could write 1 + 4 is equal to D multiplied by +13. And when we rearrange this expression to isolate and solve for D, we will find that D is simply equal to 5. In order to solve for C, we need to set X is equal to 0. So we can write 0 + 4 is equal to C multiplied by parentheses 0 minus 1, which will mean when we rearrange this expression to isolate and solve for C, we will find that C is simply equal to -4. And in order to solve for our factor B, or actually I should say for the constant B, so now we're solving for B, we need to compare the coefficients of X to the power of 1 on both sides, and when we do that, We'll be able to write 1 minus, or specifically 1 is equal to B + C, which will mean when we rearrange this expression. Or rather, before we do that, we can write that 1 is equal to -B minus 4 because we plug in a value of -4 N4C, which will mean now when we rearrange our expression to isolate sulfur B, we will find that B is simply equal to -5. And in order to solve for A, so we're solving for A now, we need to compare the coefficients of X to the power of 3 on both sides, so we could write 0 is equal to A plus D, which we can fill in our known variables, so 0 is equal to A plus 5, which means when we rearrange the expression to solve for A, A is equal to -5. So now when we substitute the constants back into our expression, we can go ahead and write that x + 4 divided by x 3 multiplied by parentheses of x minus 1 will be equal to -5 divided by x minus 5 divided by x to the 2. -4 divided by x to the power of 3. Plus 5 divided by X minus 1. So, that means when we write our integral, we will write that we will have our integral of parentheses of 5 divided by x minus 5 divided by x 2 minus 4 divided by x 3, plus 5 divided by x minus 1 in parentheses dx. So now Our fourth step is we need to integrate our integral, and we will need to integrate term by term. So let's start with the interval of -5 divided by XDX, which will be equal to -5 multiplied by the natural log of the absolute value of X plus C subscript 1. Solving for our second integral, which will be the interval of -5. And specifically, we can rewrite our expression as 5 multiplied by x to the power of -2, DX, which will be equal to -5 multiplied by, or, you know, to make it simple, just since we should know how to integrate by now. We can say that our integral, when we write it in its most simplified form, it will be equal to 5 divided by X plus C2. Fantastic. Solving for our third integral, which would be 4 multiplied by x -3 DX, which will be equal to when we write it our final answer in its most simplified form, 2 divided by x 2 plus C subscript 3. And solving for our fourth integral, which is 5, so it's going to be the integral of 5 divided by x minus 1DX, which will be equal to 5 multiplied by the natural log of the absolute value of x minus 1 plus C subscript 4. Fantastic, which will mean for our fifth and final step, we need to combine our logarithmthic terms. So we need to take -5 multiplied by the natural log of the absolute value of x plus 5 divided by x plus 2 divided by x to the 2, plus 5 multiplied by the natural log of x minus 1 in absolute value signs. Which will be equal to 5 multiplied by the natural log of the absolute value of x minus 1 divided by x plus 5 divided by x plus 2 divided by x to the power of 2 plus c where c denotes some constant variable and that's our final answer and let us make a note off to the side here. That C is equal to C1 plus C2 + C3 plus C4. Fantastic. We did it. We saw for our final answer. So, looking at the prom itself. We could safely say that our final answer for this evaluated integral is once again 5 multiplied by the natural log of the absolute value of x minus 1 divided by x plus 5 divided by x + 2 divided by x to the 2 + c. And that's it. Thank you so much for watching. Hopefully that helped, and I can't wait to see you in the next video. Bye.

Evaluate the integrals in Exercises 69–134. The integrals are listed in random order so you need to decide which integration technique to use.∫ (x + 1) / (x⁴ − x³) dx

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Evaluate the integrals in Exercises 69–134. The integrals are listed in random order so you need to decide which integration technique to use.
∫ (x + 1) / (x⁴ − x³) dx