For f(x)=int_0^x ln(1+2t^2) \ dt, we have from the FTC:
f'(x)=ln(1+2x^2)
We note the following power series:
ln(1+z) = sum _{n=1}^{\infty }(-1)^{n+1}{x^{n}}/{n} = x - x^2/2 +x^3 /3 - ... for |x|<1
So:
f'(x) = sum _{n=1}^{\infty }(-1)^{n+1}{(2x^2)^{n}}/{n}
= sum _{n=1}^{\infty }(-1)^{n+1}2^n{x^{2n}}/{n}
Integrating this:
f(x) = sum _{n=1}^{\infty }(-1)^{n+1}2^n{x^{2n+1}}/{n(2n+1)} + C
We know from the integral that f(0) = 0 implies C = 0, so:
f(x) = sum _{n=1}^{\infty }(-1)^{n+1}2^n{x^{2n+1}}/{n(2n+1)}
For convergence we apply the ratio test:
abs( ((-1)^{n+2}2^(n+1){x^{2n+3}}/{(n+1)(2n+3)} )/((-1)^{n+1}2^n{x^{2n+1}}/{n(2n+1)} ) )_(n to oo)
= abs(-2 x^2* ( n(2n+1) )/((n+1)(2n+3)) )_(n to oo)
= abs(-2 x^2* ( 2+1/n) /(2 + 5/n + 3/n^2) )_(n to oo)
implies abs(2x^2) < 1
This will definitely converge for: x in (- 1/sqrt 2, 1/sqrt 2)
