# Integrate `int 1/(5 - 4x) dx`

The integral to be evaluated is`int 1/(5-4x) dx`

Let y = 5 - 4x

`dy/dx = -4`

`dx = dy/(-4)`

Substituting the above in the original integral:

`int 1/(5 - 4x) dx`

=> `int (1/y)(1/-4)dy`

=> `(-1/4) int (1/y) dy`

=> `-(ln y)/4`

substitute y = 5 - 4x

=> `-ln(5-4x)/4 + C`

=> `-ln(5-4x)^(1/4) +C`

=> `ln(1/(5-4x)^(1/4)) + C`

The required integral is `ln(1/(5 - 4x)^(1/4)) + C`

Approved by eNotes Editorial Team

Use substitution to transform the above integral iinto an easier  to handle integral.

You should come up with the following substitution: 5-4x = u.

Differentiating the equation `5-4x = u`  yields: `-4dx = du =gt dx = -du/4`

Write the new integral:

`int (-du/4)/(u) = (-1/4)*int (du)/u`

Notice that this integral is easier to handle.

`(-1/4)*int (du)/u = (-1/4)*ln |u| + c`

Replacing u by 5-4x yields:

`int dx/(5-4x) = (-1/4)*ln |5-4x| + c`

Evaluating the integal yields:`int dx/(5-4x) = (-1/4)*ln |5-4x| + c.`

Approved by eNotes Editorial Team