Find Maclaurin series of $f(x)=\int_0^x \frac{\ln(1-t^2)}{t}\, dt$. I would like to find the Maclaurin series of a function $ f(x)=\int_0^x  \frac{\ln(1-t^2)}{t}\, dt$. Any ideas on how to approach and solve this problem?
 A: The power series for $\ln $ gives
\begin{eqnarray*}
-\ln(1-t^2)= \sum_{n=1}^{\infty} \frac{t^{2n}}{n}
\end{eqnarray*}
Now integrate term by term
\begin{eqnarray*}
\int_0^{x} \frac{\ln(1-t^2)}{t} dt = - \frac{1}{2}\sum_{n=1}^{\infty} \frac{x^{2n}}{n^2}= - \frac{1}{2} \operatorname{Li}_2(x^2).
\end{eqnarray*}
where $\operatorname{Li}_2$ is the dilogarithmic function. 
A: \begin{align}
\ln (1+t) & = t - \frac{t^2} 2 + \frac{t^3} 3 - \frac{t^4}4 + \cdots \\[10pt]
& \text{This holds if $t$ is ANY number close enough to $0$,} \\
& \text{so it holds for $-t^2$:} \\[10pt]
\ln(1-t^2) & = -t^2 - \frac{t^4} 2 -\frac{t^6} 3 - \frac{t^8} 4 - \cdots \\[10pt]
& \text{Dividing both sides by $t$, we get} \\[10pt]
\frac{\ln(1-t^2)} t & = -t - \frac{t^3} 2 -\frac{t^5} 3 - \frac{t^7} 4 - \cdots
\end{align}
Now integrate from $0$ to $x$, term by term.
Although using the formula $\displaystyle f(x) = \sum_{n=0}^\infty \frac{f^{(n)}(0)}{n!} x^n $ is the right way to procede when you first figure out the series for $e^x$ or $\cos x$ or $\log(1+x),$ etc., it is ofent better in problems like yours not to use that basic formula but to derive the result from the series for more basic functions, as above.
A: Hint: 
\begin{align*} 
f'(x) &= \frac{\ln(1-x^2)}{x} \\
f^{(2)}(x) &= \frac{2}{x^2-1} - \frac{\ln(1-x^2)}{x^2} \\
& \,\,\,\vdots
\end{align*}
you can keep differentiating and use the Maclaurin series formula,
$$ f(x) = \sum_{n=0}^\infty \frac{x^n}{n!}f^{(n)}(0) $$
