# Evauate the indefinite integral integrate of ((dx)/(4x^2-4x+5))

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### 2 Answers

You should use the following formula to evaluate the given indefinite integral such that:

`int 1/(x^2+a^2) dx = (1/a) arctan (x/a) + c`

You need to convert the given form of denominator into the form presented above using the following formula `(a-b)^2 = a^2-2ab+b^2` to complete the square in quadratic `4x^2-4x+5` such that:

`4x^2-4x = a^2-2ab =gt {(4x^2 = a^2),(4x = 2ab):}=gt` `{(a=+-2x),(4x=2*2x*b):}` `=> b = 1`

Hence, you need to add and subtract 1 such that:

`4x^2 - 4x + 1 - 1 = (2x - 1)^2 - 1`

`4x^2 - 4x + 5 = (2x - 1)^2 - 1 + 5`

`4x^2 - 4x + 5 = (2x - 1)^2 + 4`

You need to substitute `(2x - 1)^2 + 4` for `4x^2 - 4x + 5` to denominator such that:

`int 1/(4x^2 - 4x + 5)dx = int 1/((2x - 1)^2 + 4)dx`

You need to use the following substitution such that:

`2x - 1 = t => 2dx = dt => dx = (dt)/2`

Changing the variable yields:

`int 1/((2x - 1)^2 + 4)dx = int 1/(t^2 + 2^2)*(dt)/2`

`int 1/(t^2 + 2^2)*(dt)/2 = (1/2) int 1/(t^2 + 2^2) dt`

`(1/2) int 1/(t^2 + 2^2) dt = (1/2) (1/2) arctan (t/2) + c`

`(1/2) int 1/(t^2 + 2^2) dt = (1/4) arctan (t/2) + c`

Substituting back `2x - 1` for `t` yields:

`int 1/(4x^2 - 4x + 5)dx = (1/4) arctan ((2x-1)/2) + c`

**Hence, evaluating the given indefinite integral yields `int 1/(4x^2 - 4x + 5)dx = (1/4) arctan ((2x-1)/2) + c.` **

`int (dx) / (4x^2 - 4x + 5)`

To start, express the denominator of the integrand in the form a`^2 + u^2 ` . To do so, apply the completing the square method.

`=int (dx)/(4x^2-4x) +5`

`= int (dx)/(4(x^2-x)+5)`

`=int (dx)/(4(x^2-x+1/4) + 5 - 1)`

`=int (dx) /(4(x-1/2)^2 +4)`

`=1/4 int (dx) /((x-1/2)^2+1)`

`=1/4 int (dx) /(1+(x-1/2)^2)`

Now that the denominator is in the form `a^2 + u^2` , use the formula `int (du)/(a^2+u^2) = 1/a tan^(-1) u/a + C` .

`= 1/4 tan ^(-1) (x-1/2) + C`

**Hence `int (dx)/(4x^2-4x+5) = 1/4 tan^(-1) (x-1/2) + C` .**