I want to solve a problem that poses the following questions
A power series expansion for $\int\frac{\frac{1}{2}x^2-x+\ln(x+1)}{x^3}dx$ is $\sum_{1}^\infty \frac{(-1)^{n-1}}{a(n)}x^n+c$ find the value of $a(n)$.
What I did was to integrate what was given, such that
$$\int\frac{\frac{1}{2}x^2-x+\ln(x+1)}{x^3}dx=\dfrac{\left(x^2-1\right)\ln\left(x+1\right)+x}{2x^2}+c$$
In addition, using the fact that
$$x\ln(1+x)=\sum_{n=2}^\infty \frac{(-1)^n}{n-1} x^n$$
We can make the expression somewhat similar, but I don't know how I can arrive at the requested value. Any suggestions?
Best Answer
Hint:
Recall the taylor series
$$\ln(1+x) = x - \frac{1}{2}x^2 + \frac{1}{3} x^3 - \frac{1}{4}x^4 + \ldots$$
Then your integrand is
$$ \begin{align} \frac{\ln(1+x) - x + \frac{1}{2}x^2}{x^3} &= \frac{\left ( x - \frac{1}{2}x^2 + \frac{1}{3} x^3 - \frac{1}{4}x^4 + \ldots \right ) - x + \frac{1}{2}x^2}{x^3} \\ &= \frac{\frac{1}{3}x^3 - \frac{1}{4}x^4 + \ldots}{x^3} \\ &= \frac{1}{3} - \frac{1}{4}x + \ldots \end{align} $$
From here, you should integrate term-by-term (why is this allowed?) to get the answer.
I hope this helps ^_^