Question

Find `int(3x)/(x-4)^4` using substitution u=x-4. How to solve it?

Answer 1

`int \ (3x)/(x-4)^4 \ dx = -(3x-4)/(2(x-4)^3)+C`

Explanation:

We want to find:

`I = int \ (3x)/(x-4)^4 \ dx`

We can perform a simple substitution; Let

`u = x-4 => (du)/dx = 1`, and. `x=u+4`

If we perform the substitution then we get:

`I = int \ (3(u+4))/u^4 \ du`
`\ \ = 3 \ int \ (u+4)/u^4 \ du`
`\ \ = 3 \ int \ 1/u^3+4/u^4 \ du`

So we can now integrate to get:

`I = 3{-1/2 1/u^2-4/3*1/u^3}+C`
`\ \ = 3{(-3u-8)/(6u^3)}+C`
`\ \ = -{(3u+8)/(2u^3)}+C`

And restoring the substitution we get:

`I = -{(3(x-4)+8)/(2(x-4)^3)}+C`
`\ \ = -{(3x-12+8)/(2(x-4)^3)}+C`
`\ \ = -(3x-4)/(2(x-4)^3)+C`