Let $$I_n(x) = \int_0^x \frac{1}{(t^2+5)^n} dt$$, $$n = 1, 2, 3, \ldots$$. Then

We are given $$I_n(x) = \int_0^x \frac{1}{(t^2+5)^n}\, dt$$ and we need to find a recurrence relation involving $$I_5$$ and $$I_6$$.

By differentiating under the integral sign (Leibniz rule), we get $$I'_n(x) = \frac{1}{(x^2+5)^n}$$. So in particular, $$I'_5(x) = \frac{1}{(x^2+5)^5}$$.

Now we derive a reduction formula. Consider $$I_n = \int_0^x \frac{dt}{(t^2+5)^n}$$. We write the numerator as $$1 = \frac{1}{5}\bigl((t^2+5) - t^2\bigr)$$, so:

$$I_n = \frac{1}{5}\int_0^x \frac{dt}{(t^2+5)^{n-1}} - \frac{1}{5}\int_0^x \frac{t^2}{(t^2+5)^n}\,dt = \frac{1}{5}I_{n-1} - \frac{1}{5}J_n$$

where $$J_n = \int_0^x \frac{t^2}{(t^2+5)^n}\,dt$$. We evaluate $$J_n$$ using integration by parts. Let $$u = t$$ and $$dv = \frac{t}{(t^2+5)^n}\,dt$$. Then $$du = dt$$ and $$v = \frac{-1}{2(n-1)(t^2+5)^{n-1}}$$.

So $$J_n = \left[\frac{-t}{2(n-1)(t^2+5)^{n-1}}\right]_0^x + \frac{1}{2(n-1)}\int_0^x \frac{dt}{(t^2+5)^{n-1}}$$

$$= \frac{-x}{2(n-1)(x^2+5)^{n-1}} + \frac{1}{2(n-1)}I_{n-1}$$

Substituting back into our expression for $$I_n$$:

$$I_n = \frac{1}{5}I_{n-1} - \frac{1}{5}\left[\frac{-x}{2(n-1)(x^2+5)^{n-1}} + \frac{1}{2(n-1)}I_{n-1}\right]$$

$$= \frac{1}{5}I_{n-1} + \frac{x}{10(n-1)(x^2+5)^{n-1}} - \frac{1}{10(n-1)}I_{n-1}$$

$$= I_{n-1}\left(\frac{1}{5} - \frac{1}{10(n-1)}\right) + \frac{x}{10(n-1)(x^2+5)^{n-1}}$$

$$= I_{n-1}\cdot\frac{2(n-1)-1}{10(n-1)} + \frac{x}{10(n-1)} \cdot I'_{n-1}$$

since $$\frac{1}{(x^2+5)^{n-1}} = I'_{n-1}(x)$$.

$$= \frac{2n-3}{10(n-1)}I_{n-1} + \frac{x}{10(n-1)}I'_{n-1}$$

Now we put $$n = 6$$:

$$I_6 = \frac{2(6)-3}{10(5)}I_5 + \frac{x}{10(5)}I'_5 = \frac{9}{50}I_5 + \frac{x}{50}I'_5$$

Multiplying both sides by 50:

$$50I_6 = 9I_5 + xI'_5$$

Rearranging: $$50I_6 - 9I_5 = xI'_5$$

Hence, the correct answer is Option A.