Show that $\int_{0}^x \frac{1}{1+t^4} dt = x - x^5/5 + x^9/9 ..... $ where $\lvert x\rvert$ < 1

Question

Show that $\int_{0}^x \frac{1}{1+t^4} dt = x - x^5/5 + x^9/9 ..... $ where $\lvert x\rvert$ < 1

I tried expanding using binomial theorem, but I am unable to prove the series and the integral will be convergent

I am also unsure if the n'th term is written correctly. Maclaurin series is suggesting something else to me.

Please explain to me step by step.

Answer 1

The formula is not correct. Let $0<x <1$. $\frac 1 {1+t^{4}}=1-t^{4}+t^{8}+..$ and this series converges uniformly in $(0,x)$. Hence we can integrate this from $0$ to $x$ term by term and you get $\int_0^{x}\frac 1 {1+t^{4}}dt= x-\frac {x^{5}} 5+\frac {x^{9}} 9...$. Since power series expansions are unique your formula cannot be valid.