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Numerade Educator



Problem 23 Medium Difficulty

Evaluate the integral.

$ \displaystyle \int \frac{10}{(x - 1)(x^2 + 9)}\ dx $


$$\ln |x-1|-\frac{1}{2} \ln \left(x^{2}+9\right)-\frac{1}{3} \tan ^{-1}\left(\frac{x}{3}\right)+C$$


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Video Transcript

Let's evaluate the given integral here. We should do a personal fraction to composition. And looking at this quadratic equation in the denominator, this quadratic expression. Excuse me. We'd like to know whether or not this factors So here we use the discriminative, which is B squared minus four a c. So this expressions coming from engineering polynomial of degree too. So we'LL check these good minus four a. C. And it's this is less than zero than the quadratic design factor in our case. Be a zero is one see his nine? So we see b squared minus four A. C is negative and that means the quadratic dozen factor. So for the for the linear factor, we have a over X minus one and then for the quadratic factor B X plus C x squared plus nine. Now let's take this latest equation. Multiply both sides by this denominator on the bottom left, we get ten a x square plus sign plus B x plus C and then X minus one. Here we should just go ahead and factory on X square. So first, let's just simplify this right hand side. We get AKs where plus nine a plus B X squared plus C X, and then we have minus the X and then minus e. Let's go. In fact, around an X square, we have a plus B was fact right x that we have C minus B and then our constant term is nine a minus e. And from this we can get a system of equation. So looking on the left hand side, we don't see an X Square so we can write. This is zero x squared. We also don't see an X that's right, that zero X and then we still have ten. So on the left hand side, the coefficient in front of X squared to zero on the right side is a plus B, the coefficient in front of X on the right hand side of C minus B, but on the left hand side of zero. So those must be equal. And then finally, on the list, the constant term is turned on the right hand side. It's nine a minus e, so those must also be equal. This means that we have a three by three system to solve in the variables or the coefficients A, B and C let's call in the next patients off the system. We had a plus B a zero C minus B a zero. And then we also had nine. A minus C equals ten. So, Scott, and solve the system. I take this first equation, I could solve this for B take the second equation solvent for sea. And then I guess here in the last equation, weaken solve this one for sea as well. But since C equals B, we could go ahead and replace the left hand side. Here we have sea so we can replace. That would be. But then we also know that b equals negative A from the first equation. So we have negative A equals nine a minus ten. This means ten equals ten, eh? So equals one. And then from this equation over here we get B. And since E equals B C is also negative one. So we found our coefficients A B and C. Let's go ahead and plug these into the partial fraction to composition. So here I'LL need more room. So let me go to the next page to write this out. The integral now becomes a which is one over X minus one and then we had B was negative one and then see was also negative ones. And then X squared plus nine. And here's a new one. A girl. This will be easier to evaluate in the original problem. Let's go in and break this up into three into girls. So here I'LL just break this up into fractions. This's negative X x squared, plus sign and then plus negative one over X squared plus sign. So we have in a rural one over X minus one minus in general X and an expert clothes line minus and a roll one over X squared plus sign. And technically, I should be careful here. I should put my DX each time the first thing the girl we can evaluate if that minus one is bothering you just to go ahead and do it, you sub. But this integral becomes natural log Absolute value. Explain this one. So that's one of our answers. Now coming to the second integral over here with the minus sign outside here we can do a use of a SW well u equals X square plus sign. Wait, This means do you over too. His x t x so that this in a rural becomes we have a negative from this a fraction appear We're the one half from the use of and then we have General do you over you. That's negative. One half ln absolute value you and then to the back substitution replace you with X squared plus nine if you want, you can drop absolute value here because X square plus nine is positive and then we have one more In the rules ago This one also has a minus outside in the front. So they were that So for this in a rural we'LL need to tricks up Let's go ahead and take X to be three tan data the eggs to be seeking square data So this integral becomes we have the DX on top on the bottom we have nine chance Where plus nine Let me go. In fact, throughout that nine on the bottom and this is negative one over three we have seeking a top sequence where tan squared plus one is also see cancer. So those will cancel Love won over three integral of just one D data. This is negative. One over three fada And now to find data, we just take our tricks up and saw For data Tan data equals X over three. Take arc tan on both sides. That's your data. So now we've evaluated all three in the rules. Let's just go ahead and at our answers up. So I'LL go to the next place to do that. So we have natural Log X minus one. That was the first integral. Then the second one where we get to use up. We had negative one half Ellen Explorer plus nine. Then the latest tricks up gave us negative one third. It's Han Inverse X over three. And then let's not forget that constancy of integration and this is our final answer.