Question

\(\begin{array}{l} I_n\left(x\right)=\int_0^x\frac{1}{\left(t^2+5\right)^n}dt, n=1, 2, 3,\cdots\end{array}\)

Then

• A. \(50I_6, - 9I_5 = xI'_5\)
• B. \(50I_6, - 11I_5 = xI'_5\)
• C. \(50I_6, - 9I_5 = I'_5\)
• D. \(50I_6, - 11I_5 = I'_5\)

Answer 1

The Correct Option is A

To solve this problem, we need to analyze the given function and use differentiation under the integral sign. The function is given as:

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

where \(n = 1, 2, 3, \ldots\).

We're interested in finding a relation involving \(I_5\) and its derivative.

Let's differentiate \(I_5(x)\) with respect to \(x\) using the Leibniz Rule (differentiation under the integral sign):

I'_5(x) = \frac{d}{dx}\left(\int_0^x \frac{1}{(t^2 + 5)^5} \, dt\right) = \frac{1}{(x^2 + 5)^5}

For \(I_5(x)\), if we apply integration by parts to the integral definition: Let \(u = \frac{1}{(t^2 + 5)^5}\) and \(dv = dt\). Then \(du = -\frac{10t}{(t^2 + 5)^6} \, dt\) and \(v = t\).

Using integration by parts:

\int u \, dv = uv - \int v \, du

Substituting:

I_5(x) = \left. \frac{x}{(x^2 + 5)^5} \right|_0^x - \int_0^x t \cdot \left(-\frac{10t}{(t^2 + 5)^6}\right) \, dt

This gives us:

I_5(x) = \frac{x}{(x^2 + 5)^5} + 10 \int_0^x \frac{t^2}{(t^2 + 5)^6} \, dt

Notice that the additional integral can be modified to involve \(I_6(x)\):

I_6(x) = \int_0^x \frac{1}{(t^2 + 5)^6} \, dt

So, by considering the arrangement of terms we have:

x (50I_6 - 9I_5) = x \cdot I'_5

Thus, the correct answer is:

50I_6 - 9I_5 = xI'_5