Let $\mathrm{I}_{\mathrm{n}}(\mathrm{x})=\int_{0}^{\mathrm{x}} \frac{1}{\left(\mathrm{t}^{2}+5\right)^{\mathrm{n}}} \mathrm{dt}, \mathrm{n}=1,2,3, \ldots$. Then

(A) $50 \mathrm{I}_{6}-9 \mathrm{I}_{5}=\mathrm{xI}_{5}^{\prime}$

(B) $50 \mathrm{I}_{6}-11 \mathrm{I}_{5}=\mathrm{xI}_{5}^{\prime}$

(C) $50 \mathrm{I}_{6}-9 \mathrm{I}_{5}=\mathrm{I}_{5}^{\prime}$

(D) $50 \mathrm{I}_{6}-11 \mathrm{I}_{5}=\mathrm{I}_{5}^{\prime}$

JEE Main · 2022 · 28 Jul 2022 · Shift 2 · Indefinite Integrals

Let

I_{n} (x) = \int_{0}^{x} \frac{1}{( t ^{2} + 5 ) ^{n}} dt, n = 1, 2, 3, \dots

. Then

Answer: (A)

Apply integration by parts to

I_{n} (x)

I_{n} (x) = [\frac{t}{( t ^{2} + 5 ) ^{n}}]_{0}^{x} + 2 n \int_{0}^{x} \frac{t ^{2}}{( t ^{2} + 5 ) ^{n + 1}} d t

Rewrite the integral:

I_{n} (x) = \frac{x}{( x ^{2} + 5 ) ^{n}} + 2 n \int_{0}^{x} \frac{( t ^{2} + 5 ) - 5}{( t ^{2} + 5 ) ^{n + 1}} d t

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

Rearranging:

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

For

n = 5

50 I_{6} (x) - 9 I_{5} (x) = \frac{x}{( x ^{2} + 5 ) ^{5}} = I_{5}^{'} (x)

(by differentiating the definition of

I_{5}

).

Thus **

50 I_{6} - 9 I_{5} = x I_{5}^{'}

**.

JEE Main · 2022 · 28 Jul 2022 · Shift 2 · Single correct option