Question

If I_n = $\int_{}^{}\frac{dx}{(x^{2} + a^{2})^{n}}$, then

• A. I₃= $\frac{1}{4a^{2}}$ . $\frac{x}{(x^{2} + a^{2})^{2}}$+ $\frac{3}{4a^{2}}$I₂
• B. I₃= $\frac{1}{a^{2}}$.$\frac{x}{(x^{2} + a^{2})^{2}}$+ $\frac{3}{4a^{2}}$I₂
• C. I₃= $\frac{1}{4a^{2}}$ . $\frac{x}{(x^{2} + a^{2})^{2}}$– $\frac{3}{4a^{2}}$I₂
• D. None of these

Answer 1

Answer: (A)

I₃= $\frac{1}{4a^{2}}$ . $\frac{x}{(x^{2} + a^{2})^{2}}$+ $\frac{3}{4a^{2}}$I₂

Answer 2

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

(Integrating by parts)

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

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

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

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

Ž I_n+1 = $\frac{1}{2na^{2}}$ . $\frac{x}{(x^{2} + a^{2})^{n}}$ + $\frac{(2n - 1)}{2na^{2}}$ I_n Put n = 2,

I₃ = $\frac{1}{4a^{2}}$ $\frac{x}{(x^{2} + a^{2})^{2}}$ + $\frac{3}{4a^{2}}$ I₂ .