Imtiazur S. answered • 01/16/18

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Calculus - The mother of all sciences

The power series for 1/(3-x)=(1/3)∑(x/3)

^{n}where n goes from 0 to ∞ (See geometric series)^{}Integrating both sides we get

∫(1/(3-x))dx=(1/3)∫∑(x/3)

^{n}dx + CIntegrating the right hand side term by term, we get

∫(1/(3-x))dx=(1/3∫∑x

^{n+1}/((n+1)3^{n}) + CThe integral on the left is -ln(3-x) which gives us the following

ln(3-x)=-(1/3∫∑x

^{n+1}/((n+1)3n) + Cwhen x=0 the above equation becomes

ln(3)=0+C or C=ln(3)

So the power series for ln(3-x)=-(1/3∫∑xn+1/((n+1)3n) + ln(3)

So x

^{2}ln(3-x) is simply -(1/3∫∑x^{n+3}/((n+1)3n) + ln(3) (Notice that the x exponent on the Right has become n+3 where it was n+1 before. This is to account for the multiplication of that expression with x^{2}