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Lösung 2.2:3d

Aus Online Mathematik Brückenkurs 2

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Version vom 10:23, 11. Mär. 2009

Observe that the derivative of the denominator is, for the most part, equal to the numerator,

(x2+2x+2)=2x+2=2(x+1)

so we can rewrite the integral as

21x2+2x+2(x2+2x+2)dx. 

The substitution u=x2+2x+2 will therefore simplify the integral considerably,

x+1x2+2x+2dx=udu=x2+2x+2=(x2+2x+2)dx=2(x+1)dx=21udu=21lnu+C=21lnx2+2x+2+C.


Note: By completing the square

x2+2x+2=(x+1)2−12+2=(x+1)2+1

we see that x2+2x+2 is always greater than or equal to 1, so we can take away the absolute sign around the argument in ln and answer with

21ln(x2+2x+2)+C.
Von „http://wiki.math.se/wikis/2009/bridgecourse2-TU-Berlin/index.php/L%C3%B6sung_2.2:3d“