How to evaluate $int ln(1-x^9)dx$ as a power series?

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How to evaluate $int ln(1-x^9)dx$ as a power series?

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Answer 1 · 2017-03-01T23:48:14

$int \ ln(1-x^9) \ dx = -x^10/10-1/2x^19/19-1/3x^28/28-1/4x^37/37 - ....$

The general term being:

$u_n = -x^(9n+1)/(n(9n+1))$

Explanation:

A standard power series is:

$ln(1+x) = x-1/2x^2+1/3x^3-1/4x^4 + .... \ \ \$ for $|x|<1$

From this we can deduce;

$ln(1-x) = ln(1+(-x) )$
$" " = -x-1/2x^2-1/3x^3-1/4x^4 - .... \ \ \$

And so:

$ln(1-x^9) = -x^9-1/2x^18-1/3x^27-1/4x^36 - ....$

$:. int \ ln(1-x^9) \ dx= int \ -x^9-1/2x^18-1/3x^27-1/4x^36 - .... dx$
$:. " " = -x^10/10-1/2x^19/19-1/3x^28/28-1/4x^37/37 - ....$

The general term being:

$u_n = -x^(9n+1)/(n(9n+1))$

Answer 2 · 2017-05-02T02:59:54

$1/(1-x)=sum_(n=0)^oox^n$

$int1/(1-x)dx=sum_(n=0)^oointx^ndx$

$-ln(1-x)=C+sum_(n=0)^oox^(n+1)/(n+1)$

$x=0$ shows that $C=0$ :

$ln(1-x)=-sum_(n=0)^oox^(n+1)/(n+1)$

$ln(1-x^9)=-sum_(n=0)^oo(x^9)^(n+1)/(n+1)$

$color(white)(ln(1-x^9))=-sum_(n=0)^oox^(9n+9)/(n+1)$

$intln(1-x^9)dx=-sum_(n=0)^oointx^(9n+9)/(n+1)dx$

$color(white)(intln(1-x^9)dx)=C-sum_(n=0)^oox^(9n+10)/((9n+10)(n+1))$

$x=0$ shows $C=0$ :

$intln(1-x^9)dx=-sum_(n=0)^oox^(9n+10)/((9n+10)(n+1))$

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How to evaluate $int ln(1-x^9)dx$ as a power series?

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How to evaluate int ln(1-x^9)dxint ln(1-x^9)dx as a power series?

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How to evaluate $int ln(1-x^9)dx$ as a power series?