# How do you integrate int x/sqrt(3x^2-6x+17) dx using trigonometric substitution?

Jun 27, 2018

$\int \frac{x}{\sqrt{3 {x}^{2} - 6 x + 17}} \mathrm{dx} = \frac{1}{3} \sqrt{3 {x}^{2} - 6 x + 17} + \frac{1}{\sqrt{3}} \ln \left(\sqrt{3} \left\mid x - 1 \right\mid + \sqrt{3 {x}^{2} - 6 x + 17}\right) + C$

#### Explanation:

Split the rational function:

$\frac{x}{\sqrt{3 {x}^{2} - 6 x + 17}} = \frac{\left(\frac{1}{6} \left(6 x - 6\right)\right) + 1}{\sqrt{3 {x}^{2} - 6 x + 17}}$

so that using the linearity of the integral:

$\int \frac{x}{\sqrt{3 {x}^{2} - 6 x + 17}} \mathrm{dx} = \frac{1}{6} \int \frac{6 x - 6}{\sqrt{3 {x}^{2} - 6 x + 17}} \mathrm{dx} + \int \frac{\mathrm{dx}}{\sqrt{3 {x}^{2} - 6 x + 17}}$

Solve now the first integral by substituting:

$u = 3 {x}^{2} - 6 x + 17$

$\mathrm{du} = \left(6 x - 6\right) \mathrm{dx}$

so that:

$\frac{1}{6} \int \frac{6 x - 6}{\sqrt{3 {x}^{2} - 6 x + 17}} \mathrm{dx} = \frac{1}{3} \int \frac{\mathrm{du}}{2 \sqrt{u}} = \frac{1}{3} \sqrt{u} + C$

and undoing the substitution:

$\frac{1}{6} \int \frac{6 x - 6}{\sqrt{3 {x}^{2} - 6 x + 17}} \mathrm{dx} = \frac{1}{3} \sqrt{3 {x}^{2} - 6 x + 17} + C$

Solve the second integral by complete the square at the denominator:

$\int \frac{\mathrm{dx}}{\sqrt{3 {x}^{2} - 6 x + 17}} = \int \frac{\mathrm{dx}}{\sqrt{3 \left({x}^{2} - 2 x + 1\right) + 14}}$

$\int \frac{\mathrm{dx}}{\sqrt{3 {x}^{2} - 6 x + 17}} = \int \frac{\mathrm{dx}}{\sqrt{3 {\left(x - 1\right)}^{2} + 14}}$

$\int \frac{\mathrm{dx}}{\sqrt{3 {x}^{2} - 6 x + 17}} = \frac{1}{\sqrt{14}} \int \frac{\mathrm{dx}}{\sqrt{{\left(\frac{\sqrt{3} \left(x - 1\right)}{\sqrt{14}}\right)}^{2} + 1}}$

Now substitute:

$v = \frac{\sqrt{3} \left(x - 1\right)}{\sqrt{14}}$

$\mathrm{dv} = \frac{\sqrt{3}}{\sqrt{14}} \mathrm{dx}$

$\int \frac{\mathrm{dx}}{\sqrt{3 {x}^{2} - 6 x + 17}} = \frac{1}{\sqrt{3}} \int \frac{\mathrm{dv}}{\sqrt{{v}^{2} + 1}}$

and then:

$v = \tan t$ with $t \in \left(\frac{\pi}{2} , \frac{\pi}{2}\right)$

$\mathrm{dv} = {\sec}^{2} t \mathrm{dt}$

$\int \frac{\mathrm{dx}}{\sqrt{3 {x}^{2} - 6 x + 17}} = \frac{1}{\sqrt{3}} \int \frac{{\sec}^{2} t \mathrm{dt}}{\sqrt{{\tan}^{2} t + 1}}$

Using the trigonometric identity:

${\tan}^{2} t + 1 = {\sec}^{2} t$

and as $\sec t > 0$ for $t \in \left(\frac{\pi}{2} , \frac{\pi}{2}\right)$:

$\sqrt{{\tan}^{2} t + 1} = \sec t$

so:

$\int \frac{\mathrm{dx}}{\sqrt{3 {x}^{2} - 6 x + 17}} = \frac{1}{\sqrt{3}} \int \sec t \mathrm{dt}$

$\int \frac{\mathrm{dx}}{\sqrt{3 {x}^{2} - 6 x + 17}} = \frac{1}{\sqrt{3}} \ln \left\mid \sec t + \tan t \right\mid + C$

$\int \frac{\mathrm{dx}}{\sqrt{3 {x}^{2} - 6 x + 17}} = \frac{1}{\sqrt{3}} \ln \left\mid v + \sqrt{{v}^{2} + 1} \right\mid + C$

$\int \frac{\mathrm{dx}}{\sqrt{3 {x}^{2} - 6 x + 17}} = \frac{1}{\sqrt{3}} \ln \left\mid \frac{\sqrt{3} \left(x - 1\right)}{\sqrt{14}} + \sqrt{{\left(\frac{\sqrt{3} \left(x - 1\right)}{\sqrt{14}}\right)}^{2} + 1} \right\mid + C$

$\int \frac{\mathrm{dx}}{\sqrt{3 {x}^{2} - 6 x + 17}} = \frac{1}{\sqrt{3}} \ln \left(\sqrt{3} \left\mid x - 1 \right\mid + \sqrt{3 {x}^{2} - 6 x + 17}\right) + C$

Putting the results together:

$\int \frac{x}{\sqrt{3 {x}^{2} - 6 x + 17}} \mathrm{dx} = \frac{1}{3} \sqrt{3 {x}^{2} - 6 x + 17} + \frac{1}{\sqrt{3}} \ln \left(\sqrt{3} \left\mid x - 1 \right\mid + \sqrt{3 {x}^{2} - 6 x + 17}\right) + C$