JEE MAIN - Mathematics (2022 - 28th July Evening Shift - No. 9)
Let $$I_{n}(x)=\int_{0}^{x} \frac{1}{\left(t^{2}+5\right)^{n}} d t, n=1,2,3, \ldots .$$ Then :
$$50 I_{6}-9 I_{5}=x I_{5}^{\prime}$$
$$50 I_{6}-11 I_{5}=x I_{5}^{\prime}$$
$$50 I_{6}-9 I_{5}=I_{5}^{\prime}$$
$$50 I_{6}-11 I_{5}=I_{5}^{\prime}$$
Explanation
$${I_n}(x) = \int\limits_0^x {{1 \over {{{({t^2} + 5)}^n}}}dt} $$
$$ = \int\limits_0^x {{1 \over {\underbrace {{{({t^2} + 5)}^n}}_I}} \times \mathop I\limits_{II} \,dt} $$
$$ = \left. {{t \over {{{({t^2} + 5)}^n}}}} \right|_0^x - \int\limits_0^x {{{ - 2nt} \over {{{({t^2} + 5)}^{n + 1}}}} \times t\,dt} $$
$$ = {x \over {{{({x^2} + 5)}^n}}} + \int\limits_0^x {2n\left( {{{{t^2} + 5 - 5} \over {{{({t^2} + 5)}^{n + 1}}}}} \right)dt} $$
$${I_n}(x) = {x \over {{{({x^2} + 5)}^n}}} + 2n\,{I_n}(x) - 10n\,{I_{n + 1}}(x)$$
$$10n\,{I_{n + 1}}(x) - (2n - 1)\,{I_n}(x) = xI{'_n}(x)$$
For $$n = 5$$
$$50{I_6}(x) - 9{I_5}(x) = xI{'_5}(x)$$
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