Asked by Steff
this is an example from the book. I do not understand where the 1/2 comes from in (1/2u)du.
Find indefinite integral.
((.3x-.2)/(.3x^2-.4x+2))dx
u=.3x^2-.4x+1
du= 2(.3x+.2)dx
1/2u(du)= 1/2 In u + c
1/2 In(.3x^2-.4x+2) + c
Find indefinite integral.
((.3x-.2)/(.3x^2-.4x+2))dx
u=.3x^2-.4x+1
du= 2(.3x+.2)dx
1/2u(du)= 1/2 In u + c
1/2 In(.3x^2-.4x+2) + c
Answers
Answered by
MathMate
Starting with
u=.3x^2-.4x+1
we get
du/dx = 2*(.3x)-0.4
=2(0.3x-0.2)
From which we get:
2(0.3x-0.2)dx = du
or
(0.3x-0.2)dx = (1/2) du
u=.3x^2-.4x+1
we get
du/dx = 2*(.3x)-0.4
=2(0.3x-0.2)
From which we get:
2(0.3x-0.2)dx = du
or
(0.3x-0.2)dx = (1/2) du
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