Question #28fed

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Answer 1 · 2016-05-29T14:24:16

#-1/2ln(2x^2-2x+5)+C#

You wish to find:

#int(1-2x)/(2x^2-2x+5)dx#

Let #u=2x^2-2x+5#. This implies that #du=(4x-2)dx#.

Multiply the numerator of the integrand by #-2#. Balance this by multiplying the exterior of the integral by #-1//2#.

#=-1/2int(-2(1-2x))/(2x^2-2x+5)dx=-1/2int(4x-2)/(2x^2-2x+5)dx#

Substitute in #u=2x^2-2x+5# and #du=(4x-2)dx# into the integrand:

#=-1/2int(du)/u#

Note that #int(du)/u=lnabsu+C#:

#=-1/2ln(2x^2-2x+5)+C#

Note that the absolute value signs are not needed because #2x^2-2x+5>0# for all values of #x#.