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 $I_n=\int \frac{dx}{{\left(x^2+a^2\right)}^n}$ then $I_{n+1}+\frac{1-2n}{2n}\cdot \frac{1}{a^2}{I}_n$ is equal to -

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

Answer 1

Answer: (C)

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