Let I_n(x) = _0^x1(t^2+5)^n,dt, n=1,2,3,.

Question

Let $I_n(x) = \displaystyle\int_0^x\frac1{(t^2+5)^n}\,dt$, $n=1,2,3,\ldots$. Then:

Accepted Answer

Using the standard reduction formula for $I_n=\displaystyle\int_0^x\frac{dt}{(t^2+a^2)^n}$ (integration by parts on $I_{n-1}$):

$$I_{n-1} = \frac{x}{a^2(2n-2)(x^2+a^2)^{n-1}}+\frac{2n-3}{a^2(2n-2)}I_{n-2}$$

Applying with $a^2=5$ and $n=6$ (relating $I_6$ and $I_5$, and noting $I_5'(x)=\dfrac1{(x^2+5)^5}$) gives, after simplification,

$50I_6 = 9I_5+xI_5'$

Solutions · 1