# Integrating and differentiating a series

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Now, can we do something else interesting with series? Let’s try integration. Let us look at

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

Integrating that gives us

$\int\limits_0^x {\frac{{dx}}{{1 + x}}} = \ln (1 + x) = x - \frac{{{x^2}}}{2} + \frac{{{x^3}}}{3} - \frac{{{x^4}}}{4} + ... + C$

The constant of integration has to be 0. We can find that by substituting x by 0.

Integrating again gives us

$\begin{gathered} \int\limits_0^x {\ln (1 + x)dx} = \left[ {(1 + x)\ln (1 + x) - x} \right]_0^x = (1 + x)\ln (1 - x) - x \hfill \\ = \frac{{{x^2}}}{2} - \frac{{{x^3}}}{{3 \cdot 2}} + \frac{{{x^4}}}{{4 \cdot 3}} - \frac{{{x^5}}}{{5 \cdot 4}} + ... \hfill \\ \end{gathered}$

We get a constant of integration in the end, but that can easily be found to be 0.

The series for cos(x) can be found by differentiating the series for sin(x).

Up a level : Power Series
Previous page : Taylor series
Next page : LogarithmsLast modified: Mar 3, 2019 @ 15:34