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$$I_n(t)=\int_0^t\frac{1}{\left(x^5+1\right)^n}dx.$$

What is the relation between $I_{n+1}(t)$ and $I_n(t)$?

Can it be done with integration by parts?

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    $\begingroup$ for $t\rightarrow\infty$ you have simply $I_{n+1}(\infty)= (1-\frac{1}{5 n})I_n(\infty)$. $\endgroup$
    – Carlo Beenakker
    Commented Jul 15, 2021 at 17:57

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We have $$ I_{n+1}(t)=\left(1-\frac{1}{5n}\right)I_n(t)+\frac{t}{5n(t^5+1)^n}, $$ which is also compatible with Carlo Beenakker's comment above. Indeed, integrating by parts we get $$ I_n(t)=\int_0^t \frac{dx}{(x^5+1)^n}=\frac{t}{(t^5+1)^n}-\int_0^txd\left(\frac{1}{(x^5+1)^n}\right)= $$ $$ =\frac{t}{(t^5+1)^n}+\int_0^tx\frac{5nx^4}{(x^5+1)^{n+1}}dx=\frac{t}{(t^5+1)^n}+\int_0^t\frac{5n((x^5+1)-1)}{(x^5+1)^{n+1}}dx= $$ $$ =\frac{t}{(t^5+1)^n}+5nI_n(t)-5nI_{n+1}(t), $$ so $$ 5nI_{n+1}(t)=\frac{t}{(t^5+1)^n}+(5n-1)I_n(t) $$ and we get the desired relation.

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