How is $\int_0^x\sum_{k=0}^n(-t^2)^k+\frac{(-t^2)^{n+1}}{1+t^2}dt$ derived

Question

Could someone please explain, in an answer to a question here, how is

$$\int_0^x\sum_{k=0}^n(-t^2)^k+\frac{(-t^2)^{n+1}}{1+t^2}dt$$

derived from $$\int_0^x\frac{1}{1+t^2}dt$$?

Answer 1

Just subsitute $x$ in $\sum_{k=0}^nx^k=\frac{1-x^{n+1}}{1-x}$ as $-t^2$.

More detail :

$$\sum_{k=0}^n(-t^2)^k+\frac{(-t^2)^{n+1}}{1+t^2}=\frac{1-(-t^2)^{n+1}}{1+t^2}+\frac{(-t^2)^{n+1}}{1+t^2}=\frac{1}{1+t^2}.$$ I don't think it can be more explicit..