A definite integration problem , need some help understanding the solution . $$\displaystyle \int_{0}^{1}\dfrac{1-t}{1-t^{n}}\,dt=\dfrac{1}{n}\displaystyle \int_{0}^{1} \dfrac{t^{\frac{1}{n}-1}}{1-t}- \dfrac{t^{\frac{2}{n}-1}}{1-t} \,dt$$
I don't know how in the solution the expression was broken in that way, after this some beta function and di-gamma function was used which i understood

It's the substitution $u=t^n$ or $t=u^{1/n}$. Then $dt=u^{1/n-1}du/n$ and so
$$\int_0^1\frac{1-t}{1-t^n}\,dt
=\frac1n\int_0^1\frac{1-u^{1/n}}{1-u} u^{1/n-1}\,du
=\frac1n\int_0^1\frac{1-u^{2/n-1}}{1-u}\,du
-\frac1n\int_0^1\frac{1-u^{1/n-1}}{1-u}\,du.$$
The reason for writing things like
$$\int_0^1\frac{1-u^a}{1-u}\,du$$
rather than
$$\int_0^1\frac{u^a}{1-u}\,du$$
is that the former integral converges, the latter doesn't.
