$$I\_n(t)=\int\_0^t\frac{1}{\left(x^5+1\right)^n}dx.$$


What is the relation between $I\_{n+1}(t)$ and $I\_n(t)$?


Can it be done with integration by parts?



Given the integral. What's the relation between $I_{n+1}(t)$ and $I_n(t)$?
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.


