How do you evaluate the integral #int dx/(3x(x-4))#?

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Answer 1 · 2018-03-15T17:52:30

# 1/12ln|(x-4)/x|+C, or, 1/12ln|(1-4/x)|+C.#

Let, #I=intdx/{3x(x-4)}=1/3int1/{x(x-4)}dx#.

#:. I=1/3*1/4int4/{x(x-4)}dx#,

#=1/12int{x-(x-4)}/{x(x-4)}dx#,

#=1/12int[x/{x(x-4)}-(x-4)/{x(x-4)}]dx#,

#=1/12int[1/(x-4)-1/x]dx#,

#=1/12[ln|(x-4)|-ln|x|]#.

# rArr I=1/12ln|(x-4)/x|+C, or, 1/12ln|(1-4/x)|+C.#

OTHERWISE,

If we use the formula #int1/(x^2-a^2)dx=1/(2a)ln|(x-a)/(x+a)|+c#,

then, #I=1/3int1/(x^2-4x)dx=1/3int1/{(x-2)^2-2^2}dx#,

#=1/3*1/(2*2)ln|{(x-2)-2}/{(x-2)+2}|#,

#=1/12ln|(x-4)/x|+C#, as before!