$Let I_n \left(x\right)={\int }_0^x\frac{1}{{\left(t^2+5\right)}^n}dt, n=1, 2, 3, ....... Then$

$$\begin{gathered} I_n\left(x\right)={\int }_0^x\frac{dt}{{\left(t^2+5\right)}^n} \\ Applying integral by parts \\ {I}_n\left(x\right)={\left[\frac{t}{{\left(t^2+5\right)}^n}\right]}_0^x-{\int }_0^x\left(-n\right){\left(t^2+5\right)}^{-n-1}.2t^2 \\ {I}_n\left(x\right)=\frac{x}{{\left(x^2+5\right)}^n}+2n{\int }_0^x\frac{t^2}{{\left(t^2+5\right)}^{n+1}}dt \\ {I}_n\left(x\right)=\frac{x}{{\left(x^2+5\right)}^n}+2n{\int }_0^x\frac{\left(t^2+5\right)-5}{{\left(t^2+5\right)}^{n+1}}dt \\ {I}_n\left(x\right)=\frac{x}{{\left(x^2+5\right)}^n}+2n I_n\left(x\right)-10n I_{n+1}\left(x\right) \\ 10n I_{n+1}\left(x\right)+(1-2n)I_n\left(x\right)=\frac{x}{{\left(x^2+5\right)}^n} \\ Put n =5 \end{gathered}$$

$50{\mathrm{I}}_6\left(\mathrm{x}\right)-9{\mathrm{I}}_5\left(\mathrm{x}\right)=\frac{\mathrm{x}}{{\left({\mathrm{x}}^2+5\right)}^5}=\mathrm{xI}{\text{'}}_5$