Compute given integral dx/x^2+2x from 6 to 8 ?

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Answer 1 · 2018-06-03T11:44:54

#1/2 ln(16/15) #

#1/(x^2 +2x) = 1/( x(x+2) ) #

# 1/( x(x+2) ) = A/x + B/(x+2) #

#=> 1 = A(x+2) + Bx#

#x = 0 #

#=> 1 = 2A => A = 1/2 #

#x = -2 #

#=> 1 = -2B => B = -1/2 #

#1/2 int_6 ^8 1/x - 1/(x+2) dx #

#=> 1/2 [ lnx - ln|x+2| ]_6 ^8 #

#=> 1/2[ ln | x/(x+2) | ]_6 ^8 #

#=> 1/2 { ln| 8/10 | - ln|6/8 | } #

#=> 1/2 ln( 16/15 ) #

or

#ln4 - 1/2 ln 15 #