How do you evaluate #int 1/(x+7)(x^2+9)# for [-1, 3]?

1 Answer
Dharma R.
Sep 16, 2015

#-24+58ln(5/3)#

Explanation:

the given #(x^2+9)/(x+7)# can be written as #x+(-7x+9)/(x+7)#
which can be furthur reduced as #x-7+58/(x+7)#
#int_ -1^3##x-7+58/(x+7)dx#=#int_-1^3##(x^2+9)/(x+7)##dx#
#int_ -1^3##xdx-7dx+58/(x+7)dx#
#x^2/2-7x+58ln(x+7)#from #[-1 3]#
#9/2-7*3+58ln10#-#(1/2-7*(-1)+58ln6)#
#4-28+58ln(10/6)#
#-24+58ln(5/3)#
# ln stands for Natural Logarithm log_e(x)#

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