QJEEM22Maa6c3: JEE Main 2022 Mathematics - Class 12 JEEM22Maa6c3 Solution: Let Then :

${I_n}(x) = \int\limits_0^x {{1 \over {{{({t^2} + 5)}^n}}}dt}$

$= \int\limits_0^x {{1 \over {\underbrace {{{({t^2} + 5)}^n}}_I}} \times \mathop I\limits_{II} \,dt}$

$= \left. {{t \over {{{({t^2} + 5)}^n}}}} \right|_0^x - \int\limits_0^x {{{ - 2nt} \over {{{({t^2} + 5)}^{n + 1}}}} \times t\,dt}$

$= {x \over {{{({x^2} + 5)}^n}}} + \int\limits_0^x {2n\left( {{{{t^2} + 5 - 5} \over {{{({t^2} + 5)}^{n + 1}}}}} \right)dt}$

${I_n}(x) = {x \over {{{({x^2} + 5)}^n}}} + 2n\,{I_n}(x) - 10n\,{I_{n + 1}}(x)$

$10n\,{I_{n + 1}}(x) - (2n - 1)\,{I_n}(x) = xI{'_n}(x)$

For $n = 5$

$50{I_6}(x) - 9{I_5}(x) = xI{'_5}(x)$