Lösung 2.2:3e

Aus Online Mathematik Brückenkurs 2

Version vom 10:23, 11. Mär. 2009

If we differentiate the denominator in the integrand

(x2+1)=2x

we obtain almost the same expression as in the numerator; there is a constant 2 which is different. We therefore rewrite the numerator as

3x=232x=23(x2+1)

so the integral can be written as

23x2+1(x2+1)dx

and we see that the substitution u=x2+1 can be used to simplify the integral,

3xx2+1dx=udu=x2+1=(x2+1)dx=2xdx=23udu=23lnu+C=23lnx2+1+C=23ln(x2+1)+C.

In the last step, we take away the absolute sign around the argument in ln, because x2+1 is always greater than or equal to 1.