What is #int lnx^5/x^4 dx#?

1 Answer
Andy Y.
Feb 4, 2017

#intln(x^5)/x^4dx=-(3ln(x^5)+5)/(9x^3)+C#

Explanation:

If you think of the question as #intln(x^5)*1/x^4dx#, then you can use integration by parts. The formula for integration by parts is

#int f" "dg=f*g-intg" "df#.

Let #f=ln(x^5)# and #dg=1/x^4 dx#. This results in

#df=5/xdx# and #g=-1/(3x^3)#. Plugging these values into the formula for integration by parts gives

#intln(x^5)/x^4dx=ln(x^5)*(-1/(3x^3))-int(-1/(3x^3))5/xdx#
#" "=-ln(x^5)/(3x^3)+5/3int1/x^4dx#
#" "=-ln(x^5)/(3x^3)+5/3(-1/(3x^3))+C#
#" "=-ln(x^5)/(3x^3)-5/9(1/(x^3))+C#
#" "=-(3ln(x^5)+5)/(9x^3)+C#

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