# How do you integrate int x/sqrt(3x^2-6x+10) dx using trigonometric substitution?

May 21, 2018

$\int \frac{x}{\sqrt{3 {x}^{2} - 6 x + 10}} \mathrm{dx} = \frac{1}{12} \sqrt{3 {x}^{2} - 6 x + 10} + \frac{6}{\sqrt{7}} \ln | \left(\frac{1}{7} \left(3 {x}^{2} - 6 x + 10\right)\right) + \sqrt{\frac{3}{7}} \left(x - 1\right) |$

#### Explanation:

$\int \frac{x}{\sqrt{3 {x}^{2} - 6 x + 10}} \mathrm{dx}$

= $\frac{1}{6} \int \frac{6 x - 6}{\sqrt{3 {x}^{2} - 6 x + 10}} \mathrm{dx} + \frac{6}{\sqrt{3}} \int \frac{\mathrm{dx}}{\sqrt{{x}^{2} - 2 x + \frac{10}{3}}}$

For the first part let $u = 3 {x}^{2} - 6 x + 10$ ten $\mathrm{du} = \left(6 x - 6\right) \mathrm{dx}$

and integral becomes $\frac{1}{6} \int \frac{\mathrm{du}}{\sqrt{u}} = \frac{1}{6} \cdot \frac{1}{2} \cdot {u}^{\frac{1}{2}}$

= $\frac{1}{12} \sqrt{3 {x}^{2} - 6 x + 10}$

Second part can be written as

$2 \sqrt{3} \int \frac{1}{\sqrt{{\left(x - 1\right)}^{2} + \frac{7}{3}}} \mathrm{dx} =$

Now let $\sqrt{\frac{3}{7}} \left(x - 1\right) = \tan t$ then $\sqrt{\frac{3}{7}} \mathrm{dx} = {\sec}^{2} t \mathrm{dt}$ and our integral becomes

$2 \sqrt{3} \sqrt{\frac{3}{7}} \int {\sec}^{2} \frac{t}{\sec} t \mathrm{dt} = \frac{6}{\sqrt{7}} \ln | \sec t + \tan t |$

= $\frac{6}{\sqrt{7}} \ln | \sqrt{\frac{3}{7} {\left(x - 1\right)}^{2} + 1} + \sqrt{\frac{3}{7}} \left(x - 1\right) |$

= $\frac{6}{\sqrt{7}} \ln | \left(\frac{1}{7} \left(3 {x}^{2} - 6 x + 10\right)\right) + \sqrt{\frac{3}{7}} \left(x - 1\right) |$

and hence $\int \frac{x}{\sqrt{3 {x}^{2} - 6 x + 10}} \mathrm{dx}$

= $\frac{1}{12} \sqrt{3 {x}^{2} - 6 x + 10} + \frac{6}{\sqrt{7}} \ln | \left(\frac{1}{7} \left(3 {x}^{2} - 6 x + 10\right)\right) + \sqrt{\frac{3}{7}} \left(x - 1\right) |$