Có người nhờ làm, nên giải ở đây vậy 
Tính tổng của chuỗi
$$S=\sum_{n=2}^{\infty} \frac{(-2)^n}{7^n(4n-1)} \left(\frac{2x+1}{2x-4}\right)^n=\sum_{n=2}^{\infty}\frac{1}{4n-1}\left(\frac{2x+1}{7(2-x)}\right)^n \,\,(*)$$
Điều kiện chuỗi hội tụ là $-7\leq \frac{2x+1}{2-x}<7$
* Nếu $0\leq \frac{2x+1}{2-x}<7\to t^4=\frac{2x+1}{7(2-x)}$
$$(*)\Rightarrow S(t)=\sum_{n=2}^{\infty}\frac{t^{4n}}{4n-1}\Rightarrow \frac{S(t)}{t}=\sum_{n=2}^{\infty}\frac{t^{4n-1}}{4n-1}$$
$$\left(\frac{S(t)}{t}\right)'=\sum_{n=2}^{\infty}t^{4n-2}=\frac{t^6}{1-t^4}\Rightarrow\int_{0}^{t} \left(\frac{S(\tau)}{\tau}\right)'d\tau=\int_{0}^{t}\frac{\tau^6}{1-\tau^4}d\tau$$
$$\frac{S(t)}{t}=\frac{1}{12}\left ( -4t^3+3\ln\frac{1+t}{1-t}-6\arctan t \right )\Rightarrow S(t)=\frac{t}{12}\left ( -4t^3+3\ln\frac{1+t}{1-t}-6\arctan t \right )$$
Check lại đáp án.
* Nếu $-7\leq \frac{2x+1}{2-x}<0\to t^4=-\frac{2x+1}{7(2-x)}$
$$(*)\Rightarrow S(t)=\sum_{n=2}^{\infty}\frac{(-1)^nt^{4n}}{4n-1} \Rightarrow \frac{S(t)}{t}=\sum_{n=2}^{\infty}\frac{(-1)^nt^{4n-1}}{4n-1}$$
$$\left(\frac{S(t)}{t}\right)'=\sum_{n=2}^{\infty}(-1)^nt^{4n-2}=\frac{t^6}{1+t^4}\Rightarrow\int_{0}^{t} \left(\frac{S(\tau)}{\tau}\right)'d\tau=\int_{0}^{t}\frac{\tau^6}{1+\tau^4}d\tau$$
$$\frac{S(t)}{t}=\frac{1}{24}\left ( 8t^3+3\sqrt{2}\ln\frac{1+\sqrt{2}t+t^2}{1-\sqrt{2}t+t^2}+6\sqrt{2}\arctan\frac{\sqrt{2}t}{t^2-1} \right )$$
$$\Rightarrow S(t)=\frac{t}{24}\left ( 8t^3+3\sqrt{2}\ln\frac{1+\sqrt{2}t+t^2}{1-\sqrt{2}t+t^2}+6\sqrt{2}\arctan\frac{\sqrt{2}t}{t^2-1} \right )$$
Check lại đáp án.
