Question

Let I_n= $\int_{}^{}\frac{dx}{(x^{2} + a^{2})^{n}}$, where n Ī N and n > 1. If I_n and I_{n – 1} are related by the relation PI_n = $\frac{x}{(x^{2} + a^{2})^{n–1}}$ + Q I_{n – 1}. Then P and Q are respectively given by –

• A. (2n – 1)a², 2n – 3
• B. 2a² (n – 1), 2n – 3
• C. a² (n + 1), 2n + 3
• D. None of these

Answer 1

Answer: (B)

2a² (n – 1), 2n – 3

Answer 2

I_n = $\int_{}^{}\frac{dx}{(x^{2} + a^{2})^{n}}$ = $\frac{1}{(x^{2} + a^{2})^{n}}$ .

x – $\int_{}^{}\frac{( - n)2x}{(x^{2} + a^{2})^{n + 1}}$ . xdx

[Integrating by parts using 1 as second function]

\ I_n = $\frac{x}{(x^{2} + a^{2})^{n}}$ + 2n $\int_{}^{}\frac{x^{2} + a^{2} - a^{2}}{(x^{2} + a^{2})^{n + 1}}$ dx

= $\frac { x } { \left( x ^ { 2 } + a ^ { 2 } \right) ^ { n } }$ + 2n(I_n – a²I_n+1)

Ž 2na²I_n+1 = $\frac{x}{(x^{2} + a^{2})}$ + (2n – 1)I_n Replace n by n – 1,

then 2(n – 1)a² I_n = $\frac{x}{(x^{2} + a^{2})^{n - 1}}$ + (2n – 3)I_n–1