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Lösung 2.1:3d

Aus Online Mathematik Brückenkurs 2

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Version vom 10:17, 11. Mär. 2009

By dividing the two terms in the numerator by x, we can simplify each term to a form which makes it possible simply to write down the primitive functions of the integrand,

xx2+1dx=xx2+x1dx=x+x−1dx=2x2+lnx+C

where C is an arbitrary constant.


Note: Observe that 1x has a singularity at x=0, so the answers above are only primitive functions over intervals that do not contain x=0.

Von „http://wiki.math.se/wikis/2009/bridgecourse2-TU-Berlin/index.php/L%C3%B6sung_2.1:3d“