Let $I_{n}=\int_{0}^{\frac{1}{a}} \frac{1}{\left[1+(a x)^{2}\right]^{n}} d x$, where $n \in \mathbb{Z}^{+}$and $a \in \mathbb{R}^{+}$. (i) Using integration by parts, show that $I_{n}=\frac{2 n}{2 n-1} I_{n+1}-\frac{1}{a 2^{n}(2 n-1)}$.

Step 1 ：Let \(u = \frac{1}{\left[1+(a x)^{2}\right]^{n}}\) and \(dv = dx\). Then \(du = -2na(ax)^2\left[1+(a x)^{2}\right]^{-(n+1)} dx\) and \(v = x\).

Step 2 ：Using integration by parts, we have \(I_n = \int_{0}^{\frac{1}{a}} u dv = uv \Big|_0^{\frac{1}{a}} - \int_{0}^{\frac{1}{a}} v du\).

Step 3 ：Evaluating the first term, we get \(uv \Big|_0^{\frac{1}{a}} = \frac{1}{\left[1+(a \cdot \frac{1}{a})^{2}\right]^{n}} - 0 = \frac{1}{(1+a^2)^n}\).

Step 4 ：For the second term, we have \(-\int_{0}^{\frac{1}{a}} v du = 2na \int_{0}^{\frac{1}{a}} x(ax)^2\left[1+(a x)^{2}\right]^{-(n+1)} dx = 2na \int_{0}^{\frac{1}{a}} x(ax)^2 I_{n+1} dx\).

Step 5 ：Now, we can rewrite the integral as \(2na \int_{0}^{\frac{1}{a}} x(ax)^2 I_{n+1} dx = 2na \int_{0}^{1} x^3 I_{n+1} dx\).

Step 6 ：Integrating, we get \(2na \int_{0}^{1} x^3 I_{n+1} dx = 2na \cdot \frac{1}{4} I_{n+1} = \frac{na}{2} I_{n+1}\).

Step 7 ：Putting everything together, we have \(I_n = \frac{1}{(1+a^2)^n} - \frac{na}{2} I_{n+1}\).

Step 8 ：Rearranging the equation, we get \(I_n = \frac{2n}{2n-1} I_{n+1} - \frac{1}{a 2^{n}(2n-1)}\).

Step 9 ：\(\boxed{I_n = \frac{2n}{2n-1} I_{n+1} - \frac{1}{a 2^{n}(2n-1)}}\)