Solveeit Logo
Login

Question

Question: If I<sub>n</sub> = \(\int_{}^{}\frac{dx}{(x^{2} + a^{2})^{n}}\)where n Ī I and n \> I. If I<sub>n</s...

If In = ∫dx(x2+a2)n\int_{}^{}\frac{dx}{(x^{2} + a^{2})^{n}}where n Ī I and n > I. If In and In–1 are related by the relation P In = x(x2+a2)n−1\frac{x}{(x^{2} + a^{2})^{n - 1}}+ Q In–1 . Find the value of P and Q in terms of n –

A

P = 2a2 (n – 1) and Q = (2n – 3)

B

P = 2a (n – 1) and Q = (2n – 3)

C

P = 2a2 (n – 1) and Q = (4n – 3)

D

None of these

Answer

P = 2a2 (n – 1) and Q = (2n – 3)

Explanation

Solution

Q In = ∫dx(x2+a2)n\int_{}^{}\frac{dx}{(x^{2} + a^{2})^{n}}

\ In – 1 = ∫dx(x2+a2)n−1\int_{}^{}\frac{dx}{(x^{2} + a^{2})^{n - 1}}

Integrating In – 1 by parts taking unity as the second function, we have

In–1 = x(x2+a2)n−1\frac{x}{(x^{2} + a^{2})^{n - 1}} – ∫−2(n−1)x2(x2+a2)n\int_{}^{}\frac{- 2(n - 1)x^{2}}{(x^{2} + a^{2})^{n}}dx

= x(x2+a2)n–1\frac{x}{(x^{2} + a^{2})^{n–1}} + 2(n – 1) ∫(x2+a2)–a2(x2+a2)ndx\int_{}^{}\frac{(x^{2} + a^{2})–a^{2}}{(x^{2} + a^{2})^{n}}dx

= x(x2+a2)n–1\frac{x}{(x^{2} + a^{2})^{n–1}} + 2(n – 1) ∫dx(x2+a2)n–1dx\int_{}^{}\frac{dx}{(x^{2} + a^{2})^{n–1}}dx

– 2a2 (n – 1) ∫dx(x2+a2)n\int_{}^{}\frac{dx}{(x^{2} + a^{2})^{n}}

= x(x2+a2)n–1\frac{x}{(x^{2} + a^{2})^{n–1}}+ 2 (n – 1) In – 1 – 2a2 (n – 1) In

or 2a2 (n – 1) In = x(x2+a2)n−1\frac{x}{(x^{2} + a^{2})^{n - 1}} + (2n – 3) In – 1 … (1)

and given PIn = x(x2+a2)n−1\frac{x}{(x^{2} + a^{2})^{n - 1}}+ Q In–1 …(2)

Comparing (1) and (2). We get

P = 2a2 (n – 1) and Q = (2n – 3).