Evaluate: $$\int \sum_{n=0}^{\infty}(n+1)x^ndx$$ Here's how I proceded: $$\int(1 + 2x + 3x^2 + 4x^3 + \dots )dx$$ $$= x + x^2 + x^3 + x^4 + \dots + C = \frac{x}{1-x} + C$$
But the book says that the answer is: $$\frac{1}{1-x} + C$$ How?! Also, can we write the series inside the integral in a closed form, and proceed to solve further from there?
As was mentioned in the comments, $$\frac{x}{1-x}=\frac1{1-x}-1,$$ so $$\begin{align} \frac{x}{1-x}+(\text{constant})&=\frac{1}{1-x}-1+(\text{constant})\\ &=\frac{1}{1-x}+(\text{constant}), \end{align}$$ and thus your antiderivative is still correct.
EDIT:
A request was made for the closed form of the sum inside the integral. This is simple to do. Start with the sum $$\frac{1}{1-x}=\sum_{n\ge0}x^n.$$ Then differentiate both sides: $$\frac{1}{(1-x)^2}=\sum_{n\ge0}nx^{n-1}=0+\sum_{n\ge1}nx^{n-1}=\sum_{n\ge0}(n+1)x^n$$