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

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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`

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.` **