Integration (2)

$\begin{aligned} I & = \int_0^\infty \frac {dx}{(x+\sqrt{1+x^2})^8} & \small \color{#3D99F6} \text{Let }x = \tan \theta \implies dx = \sec^2 \theta \ d\theta \\ & = \int_0^\frac \pi 2 \frac {\sec^2\theta}{(\tan \theta+\sec \theta)^8}\ d \theta \\ & = \int_0^\frac \pi 2 \frac {\cos^6 \theta \ d\theta}{(\sin \theta+1)^8}\ d \theta & \small \color{#3D99F6} \text{By }\int_a^b f(x) \ dx = \int_a^b f(a+b-x) \ dx \\ & = \int_0^\frac \pi 2 \frac {\sin^6 \theta}{(\cos \theta+1)^8}\ d \theta \\ & = \int_0^\frac \pi 2 \frac {2^6\sin^6 \frac \theta 2 \cos^6 \frac \theta 2}{\left(2\cos^2 \frac \theta 2-1+1\right)^8}\ d \theta \\ & = \frac 14 \int_0^\frac \pi 2 \frac {\sin^6 \frac \theta 2}{\cos^{10} \frac \theta 2}\ d \theta \\ & = \frac 14 \int_0^\frac \pi 2 \tan^6 \frac \theta 2 \sec^4 \frac \theta 2\ d \theta & \small \color{#3D99F6} \text{Let }t = \tan \frac \theta 2 \implies dt = \frac 12 \sec^2 \frac \theta 2 \ d\theta \\ & = \frac 14 \int_0^1 t^6 (1+t^2)^2 \cdot \frac {2\ dt}{1+t^2} \\ & = \frac 12 \int_0^1 t^6 (1+t^2) \ dt \\ & = \frac 12 \left[\frac {t^7}7 + \frac {t^9}9\right]_0^1 \\ & = \frac 8{63} \end{aligned}$

Therefore, $q-p = 63-8 = \boxed{55}$ .

Refer the Result and put $n=8$

Refer Image

$Now\quad I(n)=\frac { n }{ { n }^{ 2 }-1 } \\ I(8)=\frac { 8 }{ { 8 }^{ 2 }-1 } =\frac { 8 }{ 63 } =\frac { p }{ q } \\ Now\quad q-p=63-8=55$