Taylor Series for ln(1-3x) about x = 0 | Homework Question

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Homework Statement


Determine the Taylor Series for f(x) = ln(1-3x) about x = 0

Homework Equations



ln(1+x) = \sum\fract(-1)^n^+^1 x^n /{n}

The Attempt at a Solution



ln(1-3x) = ln(1+(-3x))

ln(1+(-3x)) = \sum\fract(-1)^n^+^2 x^3^n /{n}

Is that right?
 
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The -1 is in the right place, but I'm not sure why the 3 migrated to the exponent.
 
So is it:
<br /> \sum\fract(-1)^n^+^2 3x^n /{n}<br />
 
You check it yourself by computing the first couple of terms in the Taylor series.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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