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I changed the denominator to $(x-4)^2+1$, but I'm still struggling to get it to a point where I can integrate.

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    $\begingroup$ Well, it should be $(x-4)^2+1$ $\endgroup$
    – MafPrivate
    Feb 11, 2020 at 3:47

2 Answers 2

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$$\int\dfrac1{\left(x^2 - 8x + 17\right)^{3/2}}\,\mathrm dx = \int\dfrac1{\left((x - 4)^2 + 1\right)^{3/2}}\,\mathrm dx$$

Let $x - 4 = \tan u\implies\mathrm dx = \sec^2u\,\mathrm du$.

$$\int\dfrac1{\left((x - 4)^2 + 1\right)^{3/2}}\,\mathrm dx\equiv\int\dfrac{\sec^2u}{\left(\tan^2u + 1\right)^{3/2}}\,\mathrm du\stackrel{\sec^2(u) = 1 + \tan^2(u)}=\int\dfrac1{\sec u}\,\mathrm du = \int\cos u\,\mathrm du$$

Can you take it from here?

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  • $\begingroup$ $\int cos(u)du = sin(u)$, and since $u =arctan(x-4)$, the final answer is $sin(arctan(x-4))$, correct? $\endgroup$ Feb 11, 2020 at 4:00
  • $\begingroup$ Correct. Notice that $\sin\arctan z$ can be simplified further. $\endgroup$
    – an4s
    Feb 11, 2020 at 4:02
  • $\begingroup$ You saved my life dude! $\endgroup$ Feb 11, 2020 at 4:03
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Let $u = x-4$. Then,

$$I=\int\frac{dx}{(x^2 -8x + 17)^{3/2}}=\int\frac{du}{(u^2 +1)^{3/2}}= \frac u{\sqrt{u^2+1}}+C$$

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