Let [I_n(x)=_0^x 1(t^2+5)^n d t, n=1,2,3, .] Then :

Question

Let $$I_{n}(x)=\int_{0}^{x} \frac{1}{\left(t^{2}+5\right)^{n}} d t, n=1,2,3, \ldots .$$ Then :

Accepted Answer

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

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

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

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

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

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

For $n = 5$

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

Solutions · 1