Find a closed form of the series $\sum_{n=1}^{\infty} \frac{x^n}{n^2+3n+2}$

Question

The question I've been given is this:

Using both sides of this equation: $$ \frac{x}{1-x} = \sum_{n=1}^{\infty}x^n $$

Find an expression for: $$ \sum_{n=1}^{\infty} \frac{x^n}{n^2+3n+2} $$

Any help is appreciated, thanks :)

Answer 1

Hint. You may observe that

$$ \sum_{n=1}^\infty\frac{x^n}{n}=-\ln (1-x), \quad |x|<1 $$

(this may be proved by differentiating termwise the power series).

Then $$ \frac1{n^2+3n+2}=\frac1{n+1}-\frac1{n+2} $$ gives $$ \sum_{n=1}^\infty\frac{x^n}{n^2+3n+2}=\frac1x\sum_{n=1}^\infty\frac{x^{n+1}}{n+1}-\frac1{x^2}\sum_{n=1}^\infty\frac{x^{n+2}}{n+2}, \quad 0<|x|<1. $$ Can you take it from here?

Answer 2

Hint.

$$\frac{1}{n^2+3n+2}= \frac{1}{n+1}-\frac{1}{n+2}$$

Then proceed by integration.