$I_n(x)=\int\limits_0^{ x } \frac{ dt }{\left( t ^2+5\right)^{ n }}$
Applying integral by parts
$I_n(x)=\left[\frac{t}{\left(t^2+5\right)^n}\right]_0^x-\int\limits_0^x n\left(t^2+5\right)^{-n-1} \cdot 2 t^2$
$I_n(x)=\frac{x}{\left(x^2+5\right)^n}+2 n \int\limits_0^x \frac{t^2}{\left(t^2+5\right)^{n+1}} d t$
$I_n(x)=\frac{x}{\left(x^2+5\right)^n}+2 n \int\limits_0^x \frac{\left(t^2+5\right)-5}{\left(t^2+5\right)^{n+1}} d t$
$I_n(x)=\frac{x}{\left(x^2+5\right)^n}+2 n I_n(x)-10 n I_{n+1}(x)$
$10 n I_{n+1}(x)+(1-2 n) I_n(x)=\frac{x}{\left(x^2+5\right)^n}$
$ P u t n=5$