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Aus Online Mathematik Brückenkurs 2

There is no ready made standard formula for a primitive function to our integrand, but if we expand

2−1x−2x+1dx=2−1x2+x−2x−2dx2−1x2−x−2dx

and write the last integral as

2−1x2−x1−2x0dx 

we see that the integrand consists of three terms of the type xn and we can directly write down a primitive function:

2−1x2−x1−2x0dx=3x3−2x2−2x12−1=323−222−212−3−13−2−12−21−1=38−24−4−−31−21+2=616−12−24+2+3−12=−627=−29