Animated Solution for Mathematics - Definite Integration: Let $I_n(x)={\int }_0^x\frac{dt}{(t^2+5{)}^n},n=1,2,3,\dots$. Then

Unlocking the Integral

Welcome, future engineer. Today, we are going to embark on a journey through one of the most elegant techniques in calculus: the Reduction Formula.

When you first look at the integral $I_n(x)={\int }_0^x\frac{dt}{(t^2+5{)}^n}$, it might seem daunting. The power $n$ in the denominator suggests that as $n$ increases, the integral becomes more complex.

But in mathematics, complexity is often just a mask for hidden structure. Our goal is to peel back that mask and find the relationship between $I_6$, $I_5$, and the derivative $I_5^{\prime }(x)$.

The Power of Integration by Parts

To begin, we need to break this integral down. We have a single function, but we know that Integration by Parts (IBP) is our most powerful tool for reduction.

We set $u=(t^2+5{)}^{-n}$ and $dv=dt$. Differentiating $u$ gives us $du=-n(t^2+5{)}^{-n-1}\cdot (2t)dt$, and integrating $dv$ gives us $v=t$.

Applying the IBP formula $\int udv=uv-\int vdu$, we get:

Evaluating the boundary term at $t=x$ and $t=0$, we find it simplifies beautifully to $\frac{x}{(x^2+5{)}^n}$. The integral term now looks like this: $2n{\int }_0^x\frac{t^2}{(t^2+5{)}^{n+1}}dt$.

The Algebraic Magic

Now, we face the integral $\int \frac{t^2}{(t^2+5{)}^{n+1}}dt$. This is where the magic happens. We want the numerator to look like the denominator.

By adding and subtracting $5$, we rewrite $t^2$ as $(t^2+5)-5$. This allows us to split the integral into two distinct parts:

Splitting this, we get $2n\int \frac{t^2+5}{(t^2+5{)}^{n+1}}dt-10n\int \frac{1}{(t^2+5{)}^{n+1}}dt$. The first part simplifies to $2n{I}_n$, and the second part is $-10n{I}_{n+1}$.

We have successfully expressed $I_n$ in terms of $I_n$ and $I_{n+1}$.

The Final Bridge

Rearranging our equation, we get $10n{I}_{n+1}=\frac{x}{(x^2+5{)}^n}+(2n-1)I_n$. This is our master recurrence relation.

Now, we look at the options. They involve $I_5^{\prime }(x)$. Using the Leibniz Rule, we know $I_5^{\prime }(x)=\frac{1}{(x^2+5{)}^5}$.

If we set $n=5$ in our recurrence relation, we get:

Rearranging this gives us $50I_6-9I_5=x{I}_5^{\prime }$. This matches Option (A) perfectly.

You see? By trusting the process and using the right tools, even the most intimidating integral yields to your logic. Keep practicing, keep questioning, and keep falling in love with the elegance of mathematics.