Key Concepts and Formulas

• Integration by Parts: For definite integrals, $\int_a^b u \, dv = [uv]_a^b - \int_a^b v \, du$.
• Leibniz Integral Rule (for variable upper limit): If $F(x) = \int_{a}^{x} f(t) \, dt$, then $F'(x) = f(x)$.
• Reduction Formulas: Techniques to express an integral involving an integer parameter $n$ in terms of integrals with lower or higher parameter values, often derived using integration by parts.

Step-by-Step Solution

Step 1: Define the Integral and Identify the Goal We are given the integral $I_{n}(x)=\int_{0}^{x} \frac{1}{\left(t^{2}+5\right)^{n}} d t$ for $n=1,2,3, \ldots$. Our goal is to find a relationship between $I_6(x)$, $I_5(x)$, and $I_5'(x)$. This typically involves deriving a general reduction formula for $I_n(x)$ and then substituting specific values for $n$.

Step 2: Apply Integration by Parts to $I_n(x)$ We will use integration by parts on $I_n(x)$. A strategic choice for $u$ and $dv$ is crucial. Let's set: $u = \frac{1}{(t^2+5)^n} = (t^2+5)^{-n}$ $dv = dt$

Then, we find $du$ and $v$: $du = -n(t^2+5)^{-n-1} \cdot (2t) \, dt = \frac{-2nt}{(t^2+5)^{n+1}} \, dt$ $v = \int dt = t$

Applying the integration by parts formula for definite integrals: $I_n(x) = \left[ uv \right]_0^x - \int_0^x v \, du$ $I_n(x) = \left[ t \cdot \frac{1}{(t^2+5)^n} \right]_0^x - \int_0^x t \cdot \frac{-2nt}{(t^2+5)^{n+1}} \, dt$

Step 3: Evaluate the Boundary Term and Simplify the Integral Evaluate the term $[uv]_0^x$: $\left[ \frac{t}{(t^2+5)^n} \right]_0^x = \frac{x}{(x^2+5)^n} - \frac{0}{(0^2+5)^n} = \frac{x}{(x^2+5)^n}$

Now, simplify the integral term: $- \int_0^x t \cdot \frac{-2nt}{(t^2+5)^{n+1}} \, dt = \int_0^x \frac{2nt^2}{(t^2+5)^{n+1}} \, dt$

Substituting these back into the expression for $I_n(x)$: $I_n(x) = \frac{x}{(x^2+5)^n} + \int_0^x \frac{2nt^2}{(t^2+5)^{n+1}} \, dt$

Step 4: Manipulate the Integral to Introduce $I_n(x)$ and $I_{n+1}(x)$ The integral term $\int_0^x \frac{2nt^2}{(t^2+5)^{n+1}} \, dt$ needs to be related to our defined integrals. The key is to rewrite the numerator $t^2$ in a way that includes the term $(t^2+5)$. We can do this by adding and subtracting 5: $\int_0^x \frac{2nt^2}{(t^2+5)^{n+1}} \, dt = \int_0^x \frac{2n(t^2+5-5)}{(t^2+5)^{n+1}} \, dt$ Split this into two integrals: $= \int_0^x \frac{2n(t^2+5)}{(t^2+5)^{n+1}} \, dt - \int_0^x \frac{2n \cdot 5}{(t^2+5)^{n+1}} \, dt$ $= \int_0^x \frac{2n}{(t^2+5)^n} \, dt - 10n \int_0^x \frac{1}{(t^2+5)^{n+1}} \, dt$

Recognize these integrals in terms of $I_n(x)$ and $I_{n+1}(x)$: $= 2n \, I_n(x) - 10n \, I_{n+1}(x)$

Step 5: Formulate the Reduction Formula Substitute the manipulated integral back into the equation for $I_n(x)$: $I_n(x) = \frac{x}{(x^2+5)^n} + 2n \, I_n(x) - 10n \, I_{n+1}(x)$ Rearrange the terms to isolate $I_{n+1}(x)$: $10n \, I_{n+1}(x) = \frac{x}{(x^2+5)^n} + 2n \, I_n(x) - I_n(x)$ $10n \, I_{n+1}(x) = \frac{x}{(x^2+5)^n} + (2n-1) \, I_n(x)$

Step 6: Introduce the Derivative Term The options involve $I_5'(x)$. By the Leibniz Integral Rule, the derivative of $I_n(x)$ with respect to $x$ is: $I_n'(x) = \frac{d}{dx} \int_0^x \frac{1}{(t^2+5)^n} \, dt = \frac{1}{(x^2+5)^n}$ We can use this to rewrite the term $\frac{x}{(x^2+5)^n}$ as $x \cdot \frac{1}{(x^2+5)^n} = x I_n'(x)$.

Substitute this into the reduction formula: $10n \, I_{n+1}(x) = x I_n'(x) + (2n-1) \, I_n(x)$

Step 7: Substitute $n=5$ to Match the Options We need a relation involving $I_6$ and $I_5$. This corresponds to setting $n=5$ in our general reduction formula. Let $n=5$: $10(5) \, I_{5+1}(x) = x I_5'(x) + (2(5)-1) \, I_5(x)$ $50 \, I_6(x) = x I_5'(x) + (10-1) \, I_5(x)$ $50 \, I_6(x) = x I_5'(x) + 9 \, I_5(x)$ Rearranging to match the format of the options: $50 \, I_6(x) - 9 \, I_5(x) = x I_5'(x)$

This result matches option (A).

Common Mistakes & Tips

• Incorrect Integration by Parts Setup: Choosing $u$ and $dv$ improperly can lead to more complicated integrals. For integrals of the form $\int (f(t))^n dt$, setting $u = (f(t))^n$ and $dv = dt$ is often effective.
• Algebraic Errors in Manipulation: The step of rewriting $t^2$ as $(t^2+5)-5$ is critical. Errors here will propagate through the derivation.
• Confusing $t$ and $x$: Remember that $t$ is the integration variable, while $x$ is the variable of the function $I_n(x)$ and its derivative $I_n'(x)$.

Summary We derived a reduction formula for the integral $I_n(x)$ by applying integration by parts. The process involved carefully choosing the parts for integration by parts, evaluating the boundary terms, and manipulating the resulting integral to express it in terms of $I_n(x)$ and $I_{n+1}(x)$. By utilizing the Leibniz integral rule to relate $I_n'(x)$ to the integrand evaluated at $x$, we obtained a general recurrence relation. Substituting $n=5$ into this relation yielded the specific relationship required by the problem, which is $50 I_6(x) - 9 I_5(x) = x I_5'(x)$.

The final answer is $\boxed{\text{A}}$.