take a look what Sal Khan has done with the ln(1+x^3) part of your first question
all you have to do is multiply his final series by x^2.
Of course we are not allowed to post links, so in my link, take out the space in the org part
at the start of the link
khanacademy.or g/math/ap-calculus-bc/bc-series-new/bc-10-14/v/power-series-using-integration
Find the Maclaurin series for the given function:
1) f(x) = x^2 ln(1 + x^3)
2) f(x) = x^2 / sqrt(2 + x)
3) f(x) = x*tan^(-1)(3x)
3 answers
my link worked, looks like I was able to fool the link blockage routine.
just apply the definition; find f', f", for several terms, and plug in x=0
for example, if
f = x^2 ln(1 + x^3)
f' = 2x log(1+x^3) + 3x^4/(1+x^3)
f" = 2log(1+x^3) + 9x^3(x^3+2)/(1+x^3)^2
and so on. at x=0, a lot of these are zero, so
so f(x) = x^5 + 1/2 x^8 ...
there are several handy online calculators for you to verify your results.
for example, if
f = x^2 ln(1 + x^3)
f' = 2x log(1+x^3) + 3x^4/(1+x^3)
f" = 2log(1+x^3) + 9x^3(x^3+2)/(1+x^3)^2
and so on. at x=0, a lot of these are zero, so
so f(x) = x^5 + 1/2 x^8 ...
there are several handy online calculators for you to verify your results.