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(a) Use integration by parts to show that, for any positive integer $ n $,

$$ \int \frac{dx}{(x^2 + a^2)^n}\ dx = \frac{x}{2a^2 (n - 1)(x^2 + a^2)^{n - 1}} + \frac{2n - 3}{2a^2 (n - 1)} \int \frac{dx}{(x^2 + a^2)^{n - 1}} $$

(b) Use part (a) to evaluate

$ \displaystyle \int \frac{dx}{(x^2 + 1)^2} $ and $ \displaystyle \int \frac{dx}{(x^2 + 1)^3} $

(a) $$

\int \frac{d x}{\left(x^{2}+a^{2}\right)^{n}}=\frac{x}{2 a^{2}(n-1)\left(x^{2}+a^{2}\right)^{n-1}}+\frac{2 n-3}{2 a^{2}(n-1)} \cdot \int d x\left(x^{2}+a^{2}\right)^{n-1}

$$

(b) $$

\begin{array}{l}{\int \frac{d x}{\left(x^{2}+1\right)^{2}}=\frac{x}{2\left(x^{2}+1\right)}+\frac{1}{2} \tan ^{-1} x} \\ {\int \frac{d x}{\left(x^{2}+1\right)^{3}}=\frac{x}{4\left(x^{2}+1\right)^{2}}+\frac{3 x}{8\left(x^{2}+1\right)}+\frac{3}{8} \tan ^{-1} x}\end{array}

$$

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Let's use integration by parts toe Prove the following formula and let's go ahead and denote the integral on the left. But I am so we see the end shows up on the right. Is the exponents in the denominator? So for our integration, my parts let's make this choice for you then do you? Is negative to an ex that you think the chain rule from calculus TV is the ex. So then, by using immigration my parts we know the formula. You ve todo So in our case, lets go out and write that out. Look, no, cancel those double negatives. Pull off the two one and that here are always the following trick. I'LL rewrite this as x squared plus a squared minus a squared And then I'll split it this up into two in a girls. So this is then we have X squared plus a square. Now we see that we could cancel one of these terms Run And then for the last in a girl Let's pull out that a square. We still have that too in from here. Good. And then in a girl one over x squared, plus a square. Now notice that we can rewrite these remaining two in a girls in terms of this interval right here. So this integral one over X square, plus a square to the end, is just I am. And here would you have him and plus one that will be I and plus one. So let's go to the next patient. Write this out in the first term here. This is from our UV two and to end I am minus two and a squared I and plus one that's good and soft Fur I n plus one the smiles over here. So we have this one minus to end. Yeah, no. And what's good and simplify this. Cancel all those double those minus signs and then here. Okay, so now there's new formula. This hole's for any end. So instead of using the n plus one on the left, let's just replacing within. So here, just since this works for in the end, let's replace and with and minus one that here is the left hand side just becomes I am the right hand side. Just ride it out. So this is the numerator and then on the denominator to a square. But this time and minus one going on to the next page. Simplify each term here, that's our first term. And then our second term Teo So remember, on the left this's all equal toe I end And then here on the left, I could just rewrite that first her. That's the term that we wanted from the formula. We're basically done here. The last step is it. Just replace Pull off this constant here and then replace i n minus one with the definition as thie Integral. So that's a DX X squared plus a squared all to thee and minus one power. And that's the formula that we attended on proving Now let's go ahead and start on part B. So let me go to the next page here. So we'LL plug in for actually let me just write out two Step the integral here for part B one over X squared plus one square DX. So here we see that take was one and then we have an equals two. So let's fight and use this in our formula. So we get X over to X squared plus one and then going to the second term here we plug in and equals two again. So we get one half and then in a girl, one over X squared plus one. Now, if the Senate rules bother you feel free to do it. You some here, our strengths up. Excuse me? No. And we could simplify this one half and then we have our Tana Vicks plus our constancy. Now there's two intervals in part B. This is our answer for the first animal. And we could actually use this answer from the next interval. So we're still in part be here. But the next integral this time we have X squared plus one. Cute. So using the formula from part eh? This time and equals three equals one. So we have X over We'LL have two times two x square plus once where and then the second fraction two times one times two one over one plus x clear square My notice this integral That just popped out. That's the interval we just evaluated on the last stage. So here was just plug in our answer from the other the previous part of part B And then we have three over four and then we can replace this by previous age X to X squared plus one and then we also had one half. It's an inverse X and let's got and add that constant of integration, see, and very in. And that would be our answer for the second in a girl party and that used the earlier part of the same problem. B. But it also used the formula from party, and that's our final answer.