Find alternating series that converges to $ \int_0^{1/2}x\log(1+x^3)dx $ I need to find the alternating series that converges to $ \int_0^{1/2}x\log(1+x^3)\,dx $
Here's what I did:
$$
\frac{d}{dx}[\log(1+x^3)]=\frac{1}{1+x^3}=\frac{1}{1-(-x)^2}=\sum_{n=1}^\infty(-x^3)^{n-1}=1-x^3+x^6-x^9+-...
$$
$$\begin{align}
f(x)&=x\log(1+x^3)=x\int(1-x^3+x^6-x^9+-...)\\\\
&=x\left[x-\frac{x^4}{4}+\frac{x^7}{7}-\frac{x^{10}}{10}+-...+ C\right]
\end{align}$$
$$
f(0)=0; C=0
$$
$$
f(x)=x^2-\frac{x^5}{4}+\frac{x^8}{7}-\frac{x^{11}}{10}+-...
$$
$$\begin{align}
\int_0^{1/2}x\log(1+x^3)dx&=\int_0^{1/2}(x^2-\frac{x^5}{4}+\frac{x^8}{7}-\frac{x^{11}}{10}+-...)\,dx\\\\
&=\left.\left[\frac{x^3}{3}-\frac{x^6}{6*4}+\frac{x^9}{7*9}-\frac{x^{12}}{10*12}+-...\right]\right|_0^{1/2}\\\\
&=\frac{1}{2^3(3)}-\frac{1}{2^6(6)(4)}+\frac{1}{2^9(7)(9)}-\frac{1}{2^{12}(10)(12)}+-...
\end{align}$$
Is my method correct?
 A: The logarithm is $\sum_{k=1}^\infty\frac{\left( -1\right)^{k-1}}{k}x^{3k}$. Making sure to multiply by $x$ before we integrate, the final result is $\sum_{k=1}^\infty\frac{\left( -1\right)^{k-1}}{2^{3k+2}k\left( 3k+2\right)}$.
A: Here is an efficient way forward.  We write $\log(1+x^3)=\sum_{n=1}^\infty\frac{(-1)^{n-1}x^{3n}}{n}$, which is valid for $|x|<1$.  
The series converges uniformly on $[0,1/2]$ and we may exchange the series and integration to obtain

$$\begin{align}
\int_0^{1/2}x\log(1+x^3)\,dx&=\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n}\int_0^{1/2}x^{3n+1}\,dx\\\\
&=\frac14\sum_{n=1}^\infty \frac{(-1)^{n-1}}{n(3n+2)2^{3n}}
\end{align}$$


It might be of interest to note that the integrand has an antiderivative in terms of elementary functions, which can be obtained by integration by parts and partial fraction expansion (SEE HERE).  Hence the integral can be evaluated over any interval in closed form.  Here, we find that 
$$\int_0^{1/2}x\log(1+x^3)\,dx=\frac{\sqrt 3 \pi}{12}-\frac{3\log(4e)}{16}$$
A: I would do it this way:
Using
$\ln(1+z)
=\sum_{k=1}^{\infty} \dfrac{(-1)^{k-1}x^k}{k}
$,
which converges for your range,
$\begin{array}\\
\int_0^{1/2}x\ln(1+x^3)dx
&=\int_0^{1/2}x\left(\sum_{k=1}^{\infty} (-1)^{k-1}\dfrac{x^{3k}}{k}\right)dx\\
&=\sum_{k=1}^{\infty} \dfrac{(-1)^{k-1}}{k}\int_0^{1/2}x^{3k+1}dx\\
&=\sum_{k=1}^{\infty} \dfrac{(-1)^{k-1}}{k}\dfrac{x^{3k+2}}{3k+2}\big|_0^{1/2}\\
&=\sum_{k=1}^{\infty} \dfrac{(-1)^{k-1}}{k(3k+2)2^{3k+2}}\\
&=\dfrac1{5\cdot 2^5}-\dfrac1{2\cdot8\cdot 2^8}+...\\
\end{array}
$
More generally,
$\begin{array}\\
\int x^a\ln(1+x^b)dx
&=\int x^a\left(\sum_{k=1}^{\infty} (-1)^{k-1}\dfrac{x^{bk}}{k}\right)dx\\
&=\sum_{k=1}^{\infty} \dfrac{(-1)^{k-1}}{k}\int x^{a+bk}dx\\
&=\sum_{k=1}^{\infty} \dfrac{-1)^{k-1}x^{a+1+bk}}{k(a+1+bk)}\\
\end{array}
$
