# `int x/sqrt(x^2+6x+12) dx` Complete the square and find the indefinite integral

Recall indefinite integral follows `int f(x) dx = F(x)+C`

where:

`f(x)` as the integrand

`F(x)` as the antiderivative of `f(x)`

`C` as the constant of integration.

To evaluate the given integral:  `int x/sqrt(x^2+6x+12)dx` , we  may apply completing the square at the trinomial: `x^2+6x+12` .

Completing the square:

`x^2+6x+12` is in a...

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Recall indefinite integral follows `int f(x) dx = F(x)+C`

where:

`f(x)` as the integrand

`F(x)` as the antiderivative of `f(x)`

`C` as the constant of integration.

To evaluate the given integral:  `int x/sqrt(x^2+6x+12)dx` , we  may apply completing the square at the trinomial: `x^2+6x+12` .

Completing the square:

`x^2+6x+12` is in a form of `ax^2 +bx+c`

where:

`a =1`

`b =6`

`c= 12`

To complete square ,we add and subtract `(-b/(2a))^2` :

With `a=1 ` and `b = 6` then:

`(-b/(2a))^2 =(-6/(2*1))^2 = 9`

Then `x^2+6x+12` becomes:

`x^2+6x+ 12 +9-9`

`(x^2+6x+9) + 12 -9`

`(x+3)^2 +3`

Applying `x^2 +6x +12 =(x+3)^2 + 3` in the given integral, we get:

`int x/sqrt(x^2+6x+12)dx=int x/sqrt((x+3)^2 + 3)dx`

We may apply u-substitution by letting:  `u = x+3` or` x =u-3` then `du = dx` .

The integral becomes:

`int x/sqrt((x+3)^2 + 3)dx =int (u-3)/sqrt(u^2 + 3)du`

Apply the basic integration property: `int (u-v) dx = int (u) dx - int (v) dx` .

`int (u-3)/sqrt(u^2 + 3)du =int u/sqrt(u^2 + 3)du -int 3/sqrt(u^2 + 3)du`

For the integral of  `int u/sqrt(u^2 + 3)du` , we may apply formula from integration table: `int u/sqrt(u^2+-a^2) du = sqrt(u^2+-a^2) +C`

Take note we have + sign inside the root then we follow: `int u/sqrt(u^2+a^2) du = sqrt(u^2+a^2) +C` .

`int u/sqrt(u^2 + 3)du=sqrt(u^2+3) `

For the integral of `int 3/sqrt(u^2 + 3)du` , we use the basic integration property: `int cf(x)dx = c int f(x) dx.`

`int 3/sqrt(u^2 + 3)du = 3int 1/sqrt(u^2 + 3)du`

From integration table, we may apply the formula for rational function with roots:

`int 1/sqrt(x^2+-a^2)dx = ln|x+sqrt(x^2+-a^2)| +C`

With just `(+)` inside the root, we follow:`int 1/sqrt(x^2+a^2)dx = ln|x+sqrt(x^2+a^2)| ` +C.

`3int 1/sqrt(u^2 + 3)du=ln|u+sqrt(u^2+3)|`

Combining the results, we get:

`int (u-3)/sqrt(u^2 + 3)du =sqrt(u^2+3) -ln|u+sqrt(u^2+3)| +C`

Plug-in `u = x+3` on `sqrt(u^2+3) -ln|u+sqrt(u^2+3)| +C` , we get the indefinite integral as:

`int x/sqrt(x^2+6x+12)dx =sqrt((x+3)^2+3) -ln|x+3+sqrt((x+3)^2+3)| +C`

Recall: `(x+3)^2+3 =x^2+6x+12` then the indefinite integral can also be expressed as:

`int x/sqrt(x^2+6x+12)dx =sqrt(x^2+6x+12) -ln|x+3+sqrt(x^2+6x+12)| +C`

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