# How do you integrate int x/sqrt(4x^2+4x+-24)dx using trigonometric substitution?

Aug 5, 2018

$I = \frac{1}{2} \sqrt{{x}^{2} + x \pm 6} - \frac{1}{4} \ln | x + \frac{1}{2} + \sqrt{{x}^{2} + x \pm 6} | + C$

#### Explanation:

Without trigonometric substitution :

Here ,

$I = \int \frac{x}{\sqrt{4 {x}^{2} + 4 x \pm 24}} \mathrm{dx}$

$= \frac{1}{2} \int \frac{x}{\sqrt{{x}^{2} + x \pm 6}} \mathrm{dx}$

$= \frac{1}{4} \int \frac{2 x}{\sqrt{{x}^{2} + x \pm 6}} \mathrm{dx}$

$= \frac{1}{4} \int \frac{2 x + 1 - 1}{\sqrt{{x}^{2} + x \pm 6}} \mathrm{dx}$

$= \frac{1}{4} \int \frac{2 x + 1}{\sqrt{{x}^{2} + x \pm 6}} \mathrm{dx} - \frac{1}{4} \int \frac{1}{\sqrt{{x}^{2} + x \pm 6}} \mathrm{dx}$

$= \frac{1}{4} \int \frac{d \left({x}^{2} + x \pm 6\right)}{\sqrt{{x}^{2} + x \pm 6}} - \frac{1}{4} {I}_{1}$

$= \frac{1}{4} \cdot 2 \sqrt{{x}^{2} + x \pm 6} - \frac{1}{4} {I}_{1}$

$= \frac{1}{2} \sqrt{{x}^{2} + x \pm 6} - \frac{1}{4} {I}_{1.} \ldots \ldots \ldots \ldots \ldots \ldots . . \to \left(A\right)$

Now ,

${I}_{1} = \int \frac{1}{\sqrt{{x}^{2} + x \pm 6}} \mathrm{dx}$

Now ,

$\left(i\right) {x}^{2} + x + 6 = {x}^{2} + x + \frac{1}{4} + \frac{23}{4} = {\left(x + \frac{1}{2}\right)}^{2} + \frac{23}{4}$

$\left(i i\right) {x}^{2} + x - 6 = {x}^{2} + x + \frac{1}{4} - \frac{25}{4} = {\left(x + \frac{1}{2}\right)}^{2} - \frac{25}{4}$

So ,

${I}_{1} = \int \frac{1}{\sqrt{{\left(x + \frac{1}{2}\right)}^{2} + K}} \mathrm{dx}$ ,where ,$K = \frac{23}{4} \mathmr{and} K = - \frac{25}{4}$

$\therefore {I}_{1} = \ln | \left(x + \frac{1}{2}\right) + \sqrt{{\left(x + \frac{1}{2}\right)}^{2} + K} | + c '$

$\therefore {I}_{1} = \ln | \left(x + \frac{1}{2}\right) + \sqrt{{x}^{2} + x + \frac{1}{4} + K} | + c '$

substitute $K = \frac{23}{4} \mathmr{and} K = - \frac{25}{4}$

$\therefore {I}_{1} = \ln | x + \frac{1}{2} + \sqrt{{x}^{2} + x \pm 6} | + c '$

From $\left(A\right)$ we get

$I = \frac{1}{2} \sqrt{{x}^{2} + x \pm 6} - \frac{1}{4} \ln | x + \frac{1}{2} + \sqrt{{x}^{2} + x \pm 6} | + C$

Note :

color(red)(int(f'(x))/sqrt(f(x))dx=2sqrt(f(x))+c

OR

color(red)(int(d[f(x)])/sqrt(f(x))=2sqrt(f(x))+c