Determine a Maclaurin series for

Question

a). ∫((e^x-1)/x^2) dx
b). ln⁡(3+4x)

Answer 1

I'll get you started on (b)
f = ln(3+4x)
f' = 4/(3+4x)
f" = -16/(3+4x)^2
f'" = 128/(3+4x)^3
...
so now just evaluate all these at x=0 and you have
ln(3+4x) = ln3 + 4/3 x - 8/9 x^2 + 64/81 x^3 - ...