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

Question

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

Answer 1

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

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?

Answer 2

Let $u = x-4$. Then,

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