Let I _ n (x) = ∫ _ 0 ^ x 1 ( t ^ 2 + 5) ^ n dt, n = 1, 2, 3, . Then

Given,

${I}_{n}\left(x\right) = {∫}_{0}^{x}\dfrac{dt}{{\left({t}^{2} + 5\right)}^{n}}$

Applying integration by parts we get,

$⇒ {I}_{n}\left(x\right) = {\left[\dfrac{t}{{\left({t}^{2} + 5\right)}^{n}}\right]}_{0}^{x} - {∫}_{0}^{x}n{\left({t}^{2} + 5\right)}^{- n - 1} \cdot 2{t}^{2}$

$⇒ {I}_{n}\left(x\right) = \dfrac{x}{{\left({x}^{2} + 5\right)}^{n}} + 2n{∫}_{0}^{x}\dfrac{{t}^{2}}{{\left({t}^{2} + 5\right)}^{n + 1}}dt$

$⇒ {I}_{n}\left(x\right) = \dfrac{x}{{\left({x}^{2} + 5\right)}^{n}} + 2n{∫}_{0}^{x}\dfrac{\left({t}^{2} + 5\right) - 5}{{\left({t}^{2} + 5\right)}^{n + 1}}dt$

$⇒ {I}_{n}\left(x\right) = \dfrac{x}{{\left({x}^{2} + 5\right)}^{n}} + 2n{∫}_{0}^{x}\dfrac{dt}{{\left({t}^{2} + 5\right)}^{n}} - 10n{∫}_{0}^{x}\dfrac{dt}{{\left({t}^{2} + 5\right)}^{n + 1}}$

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

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

$⇒ 10n{I}_{n + 1}\left(x\right) + \left(1 - 2n\right){I}_{n}\left(x\right) = x{I}^{'}n$ {where ${I}^{'}n = \dfrac{1}{{\left({x}^{2} + 5\right)}^{n}}$ we get by differentiating ${I}_{n}\left(x\right) = {∫}_{0}^{x}\dfrac{dt}{{\left({t}^{2} + 5\right)}^{n}}$ }

Now put $n = 5$

We get, $50{I}_{6} - 9{I}_{5} = x{I}_{5}^{'}$