Question

Integrals of class of functions following a definite pattern can be found by the method of reduction and recursion. Reduction formulas make it possible to reduce an integral dependent on the index $n >0$, called the order of the integral, to an integral of the same type with a smaller index. Integration by parts helps us to derive reduction formulas. (Add a constant in each question)
If $I_n=\int \frac{d x}{\left(x^2+a^2\right)^n}$ then $I_{n+1}+\frac{1-2 n}{2 n} \cdot \frac{1}{a^2} I_n$ is equal to -

• A. $\frac{x}{\left(x^2+a^2\right)^n}$
• B. $\frac{1}{2 n a^2} \frac{1}{\left(x^2+a^2\right)^{n-1}}$
• C. $\frac{1}{2 n a^2} \cdot \frac{x}{\left(x^2+a^2\right)^n}$
• D. $\frac{1}{2 n a^2} \cdot \frac{1}{\left(x^2+a^2\right)}$

Answer 1

Answer: (C)

$I_n=\frac{x}{\left(x^2+a^2\right)^n}+2 n \int \frac{x^2}{\left(x^2+a^2\right)^{n+1}} d x$
$=\frac{x}{\left(x^2+a^2\right)^n}+2 n\left[\int\left(\frac{x^2+a^2}{\left(x^2+a^2\right)^{n+1}}-\frac{a^2}{\left(x^2+a^2\right)^{n+1}}\right) d x\right] $
$=\frac{x}{\left(x^2+a^2\right)^n}+2 n \int \frac{1}{\left(x^2+a^2\right)^n} d x$
$-2 n a^2 \int \frac{1}{\left(x^2+a^2\right)^{n+1}} d x $
$=\frac{x}{\left(x^2+a^2\right)^n}+2 n\left(I_n-a^2 I_{n+1}\right)$
Whence $I_{n+1}+\frac{1-2 n}{2 n} \frac{1}{a^2} I_n=\frac{1}{2 n a^2} \cdot \frac{x}{\left(x^2+a^2\right)^n}$