Find the indefinite integral $\int \left(k+2\right)^{-1}\sum_{k=0}^{\infty}\left(\frac{t^2}{1+t^2}\right)^{k+2}dt$

Question

I need to find the indefinite integral \begin{align}I&=\int \sum_{k=0}^{\infty}\left(k+2\right)^{-1}\left(\frac{t^2}{1+t^2}\right)^{k+2}dt.\end{align} It is the result of removing the first two terms from a logarithm Taylor series (they were integrable). Now I need to evaluate this integral/summation. Any ideas? (And this isn't a homework problem, by the way.)

Answer 1

Hint: That is just: $$ \int\left(-\frac{t^2}{1+t^2}+\log(1+t^2)\right)\,dt = C-3t+3\arctan(t)+t\log(1+t^2).$$