We have

`1/(1-x)=(1-x)^(-1)=sum_(n=o)^oox^n`

Thus

`3/(2+x^3)=3/(2(1+x^3/2))=(3/2)(1-(-x^3/2))^(-1)=sum_(n=0)^oo(-x^3/2)^n`

`=(3/2)(1-x^3/2+x^6/4-x^9/8+……..)`

Thus

`a_n=(-x^3/2)^n`

`a_(n+1)=(-x^3/2)^(n+1)`

`So`

`lim_(n->oo)|a_(n+1)/a_n|=lim_(n->oo)|-x^3/2|=|x^3/2|`

Thus series will converge if `|x^3/2|<1` `=> -2^(1/3)<x<2^(1/3)`

Thus radius of the convergence of the series is `-2^(1/3)<x<2^(1/3)`

2. `int_0^1(3x)/(1+x^3)dx=int_0^1(xsum_(n=0)^ooa_n)dx`

`=sum_(n=0)^ooint_0^1(3/2)x(-x^3/2)^ndx`

`=(3/2)sum_(n=0)^ooint_0^1(-1/2)^nx^(3n+1)dx`

`=(3/2)sum_(n=0)^oo(-1/2)^n{x^(3n+2)/(3n+2)}_0^1`

`=(3/2)sum_(n=0)^oo(-1/2)^n(1/(3n+2))`

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