Question

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

Answer 1

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

Explanation:

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!