Let I n x = ∫ 0 x 1 t 2 + 5 n d t , n = 1 , 2 , 3 , … . Then

Given,

$I_n\left(x\right)={\int }_0^x\frac{dt}{{\left(t^2+5\right)}^n}$

Applying integration by parts we get,

$\Rightarrow I_n\left(x\right)={\left[\frac{t}{{\left(t^2+5\right)}^n}\right]}_0^x-{\int }_0^{x}n{\left(t^2+5\right)}^{-n-1}·2t^2$

$\Rightarrow I_n\left(x\right)=\frac{x}{{\left(x^2+5\right)}^n}+2n{\int }_0^x\frac{t^2}{{\left(t^2+5\right)}^{n+1}}dt$

$\Rightarrow I_n\left(x\right)=\frac{x}{{\left(x^2+5\right)}^n}+2n{\int }_0^x\frac{\left(t^2+5\right)-5}{{\left(t^2+5\right)}^{n+1}}dt$

$\Rightarrow I_n\left(x\right)=\frac{x}{{\left(x^2+5\right)}^n}+2n{\int }_0^x\frac{dt}{{\left(t^2+5\right)}^n}-10n{\int }_0^x\frac{dt}{{\left(t^2+5\right)}^{n+1}}$

$\Rightarrow I_n\left(x\right)=\frac{x}{{\left(x^2+5\right)}^n}+2n{I}_n\left(x\right)-10n{I}_{n+1}\left(x\right)$

$\Rightarrow 10n{I}_{n+1}\left(x\right)+\left(1-2n\right)I_n\left(x\right)=\frac{x}{{\left(x^2+5\right)}^n}$

$\Rightarrow 10n{I}_{n+1}\left(x\right)+\left(1-2n\right)I_n\left(x\right)=x{I^{\text{'}}}_n$ {where ${I^{\text{'}}}_n=\frac{1}{{\left(x^2+5\right)}^n}$ we get by differentiating $I_n\left(x\right)={\int }_0^x\frac{dt}{{\left(t^2+5\right)}^n}$ }

Now put $n=5$

We get, $50I_6-9I_5=x{I}_5^{\text{'}}$