Evaluate $\int x^2\ln (1+x)dx$ as a power series.

Question

I do not understand the justification for omitting the constant in step 4. $x^2($series$+C)=$new series$+Cx^2$. You end up getting a $Cx^2$ which is not a constant.

Answer 1

You have: $\frac {d}{dx} ln (1+x) = \frac {1}{1+x} = \sum_\limits{n=0}^{\infty} (-x)^n$

$\displaystyle\int_0^x \frac {d}{dx} ln (1+x) = \ln (1+x) + \ln 1 = \ln (1+x)$

When we integrate the power series similarly we evaluate at $x$ and at $0.$ But the power-series evaluated at $0$ equals $0.$