How do you integrate #67/x - 5sqrt(x)#?

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Answer 1 · 2015-02-01T20:59:49

You can manipulate your function to get:

#int[67/x-5x^(1/2)]dx=#

that can be written as:

#int67/xdx-int5x^(1/2)dx=#

Now you can use the fact that #intx^ndx=x^(n+1)/(n+1)+c# and #int1/xdx=ln(x)+c#

To get:

#=67ln(x)-5x^(3/2)/(3/2)+c#
#=67ln(x)-10xsqrt(x)/3+c#