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Evaluate the integral.

$ \displaystyle \int \frac{dt}{(t^2 - 1)^2} $

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Calculus 2 / BC

Chapter 7

Techniques of Integration

Section 4

Integration of Rational Functions by Partial Fractions

Integration Techniques

Campbell University

Oregon State University

Baylor University

University of Michigan - Ann Arbor

Lectures

01:53

In mathematics, integration is one of the two main operations in calculus, with its inverse, differentiation, being the other. Given a function of a real variable, an antiderivative, integral, or integrand is the function's derivative, with respect to the variable of interest. The integrals of a function are the components of its antiderivative. The definite integral of a function from a to b is the area of the region in the xy-plane that lies between the graph of the function and the x-axis, above the x-axis, or below the x-axis. The indefinite integral of a function is an antiderivative of the function, and can be used to find the original function when given the derivative. The definite integral of a function is a single-valued function on a given interval. It can be computed by evaluating the definite integral of a function at every x in the domain of the function, then adding the results together.

27:53

In mathematics, a technique is a method or formula for solving a problem. Techniques are often used in mathematics, physics, economics, and computer science.

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09:55

Evaluate the integral.…

01:32

Evaluate the integrals.

01:18

01:02

Evaluate the definite inte…

03:06

00:39

Let's evaluate the definite rule. Here we have a fraction and into Rend. Let's go ahead and do partial fraction to composition here. The first thing we should check is if this quadratic factors and it does, this is T minus one. The students he plus one square and then t minus one squared. So this tells us that the partial fraction should look like a over T plus one plus B over T plus once where this is what the textbook calls case, too repeated linear factors See over T minus one D over T minus one squared. Now it's good. And multiply both sides of this equation over here By this denominator on the left. When we do so, we have one. Then we have a P plus one T minus one square for B. We have t minus one square for sea. We have t minus one, then t plus one squared. And for D, we just have t plus one squared. So on the right hand side, let's just go ahead and multiple pile on DH, simplify as much as we can. So go ahead and foil this. Multiply out this quadratic and then we'LL supply to t plus one. Then the coefficient. We have a and then we have a cubic after simplifying. Then for be we just have got and foil that for. See, we have some work to do here. This is Cube T Square, minus T minus one. And finally for D T's Claire to T and a one now on the right hand side. Let's go. In fact, there are pretty cute. We have a policy, then was full of T squared. We have minus a a plus B plus C plus the and then pull out a T minus a minus two B minus E plus two D. And then finally the constant term A plus B minus C plus the now the left hand side. This is all equal to one. If we want, we could write the left hand side is zero t cubed plus zero t square plus zero t plus one, and then we see that we must have a plus. C equals zero because those are the co efficient of T. Q. Similarly, the coefficient in front of T Square those have to be equal. The coefficients in front of T have to be equal So we have another equation. This is all people to zero, and then we have. The constants are among the left, is one on the right hand side. There's a constant sir, so those must be equal. And this gives us a four by four system and the coefficients a through D. So it's right out these equations and solve the system. On the next page, we had a policy equal zero minus a plus. B plus C plus three equals zero minus minus a minus two B minus. C plus two B equals zero and then a plus B minus C plus the equals one. From this first equation, we can go ahead and saw that a equals might see if we want to go in and sell for a and then we can go ahead and plug this into each of the equations for a So when we plug it into the second equation, that becomes a C. So we have a B plus two C plus the zero when we plug it into this equation over here, we have a negative, eh? So that's a plus. He that'll cancel with the other seat. So we just have minus two B plus two B equals zero. And then over here we have a minus C another minus c So B minus two C plus d equals one. So from the second equation here we see that we have b equals D so we can go ahead and plug in this value of be into the other two equations here. So plugging it to the one on the left we have to see plus two d equals zero. So we cc equals negativity and then plugging it in Over here we have to. The mine is to see equals one. So at this point, we have be healthy and then c equals negative d equals negative B. So this also tells us that because a equals negative c we must also have is equal to be indeed. And then at this point, we could end this last equation. Over here, we know that C equals negative D. So this gives us forty equals one d equals one over four. Therefore, a beer also equals one over four and then we have C equals negative one over four. So we have our coefficients A, B, C and D. Now we can go back to our partial fraction to composition that we wrote plug in these values for a, B, C and D. Let's go to the next occasion. Do that. We'LL have the integral now plugging in our car coefficients one over fourth oversee plus one, one over fourth over T plus one square minus one over fourth. We're t minus one because we did have a negative one over four for the value of C, and then D was also one over four. At this point, we can go ahead. And if this plus one and minus one or bothering you on the tee, you could go ahead and do your substitution. For the first two, you could do t plus one. And then for the last two, you could do T minus one. We cannot weaken, plugged the one over four and then when we integrate, the first term should give us natural log t plus one in the absolute value. Sell me this one fourth is factored out of the whole expression. And then when we use the power rule, might this one over t plus one and then for the third mightiest natural log absolute value T minus one and then for the very last term, once again using the power rule minus one over T minus one and then plus our constancy of integration. And at this point, we could go ahead and simplify this a little bit. For example, we could combine these logs. So if you like using your lot properties, you could have t plus one over T minus one and then minus one over t plus one minus one over T minus one plus he and there's our answer. But let me circle this answer, because this is how the book rights and there's our answer.

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