$\int\frac{1}{(x^2 -8x + 17)^{3/2}}dx$

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

• Well, it should be $(x-4)^2+1$ Feb 11, 2020 at 3:47

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

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?

• $\int cos(u)du = sin(u)$, and since $u =arctan(x-4)$, the final answer is $sin(arctan(x-4))$, correct?
– user738554
Feb 11, 2020 at 4:00
• Correct. Notice that $\sin\arctan z$ can be simplified further.
– an4s
Feb 11, 2020 at 4:02
• You saved my life dude!
– user738554
Feb 11, 2020 at 4:03

Let $$u = x-4$$. Then,

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