Compute the following integral.
![\[ \int \frac{\sqrt{3-x^2}}{x} \, dx. \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-bb55d87f403f6e4fcf782f2584ccf19b_l3.png "Rendered by QuickLaTeX.com")
First, we multiply the numerator and denominator by 
and do some rearranging to get a friendlier integral,
![\begin{align*} \int \frac{\sqrt{3-x^2}}{x} \, dx &= \int \frac{3-x^2}{x \sqrt{3-x^2}} \, dx \\[9pt] &= - \int \frac{x}{\sqrt{3-x^2}} \, dx + 3 \int \frac{1}{x \sqrt{3-x^2}} \, dx. \end{align*}](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-4d2bd334c12cb7b54c2c6b9649e8d3f7_l3.png "Rendered by QuickLaTeX.com")
For the integral on the left we make the substitution 
, 
and obtain
![\[ -\int \frac{x}{\sqrt{3-x^2}} \, dx = \frac{1}{2} \int \frac{du}{\sqrt{u}} = \sqrt{u} = \sqrt{3-x^2}. \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-bbaad78d4ebecda46d86a55ae9e10d9a_l3.png "Rendered by QuickLaTeX.com")
Then, for the integral on the right we make the substitution 
, 
. This gives us
![\[ x = \sqrt{3-u^2}, \qquad dx = -\frac{\sqrt{3-x^2}}{x}. \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-993b3d3ae74a5f20434b6ebcc4798055_l3.png "Rendered by QuickLaTeX.com")
Therefore for the integral on the right we have
![\begin{align*} 3 \int \frac{1}{x\sqrt{3-x^2}} \, dx &= 3 \int \frac{1}{3-u^2} \, du \\[9pt] &= -3 \int \frac{1}{(\sqrt{3} + u)(\sqrt{3} - u)} \, du. \end{align*}](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-accccc983947a7d4e4f47ec8531e25a8_l3.png "Rendered by QuickLaTeX.com")
Now, we have to use partial fractions on the integrand,
![\[ \frac{1}{(\sqrt{3} + u)(\sqrt{3} - u)} = \frac{A}{\sqrt{3} + u} + \frac{B}{\sqrt{3} - u}. \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-81b53ab5677119e321e409123c2e31b9_l3.png "Rendered by QuickLaTeX.com")
This gives us the equation
![\[ A(\sqrt{3} - u) + B(\sqrt{3} + u) = 1 \quad \implies \quad A = B = \frac{1}{2\sqrt{3}}. \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-e90a4129c9099f4d882ff7201f6b1fc7_l3.png "Rendered by QuickLaTeX.com")
So,
![\begin{align*} -3 \int \frac{1}{(\sqrt{3} + u)(\sqrt{3} - u)} \, du &= -3 \int \left(\frac{1}{2 \sqrt{3} (\sqrt{3} + u)} - \frac{1}{2 \sqrt{3} (\sqrt{3} - u)} \right) \, du \\[9pt] &= -\frac{\sqrt{3}}{2} \left( \frac{1}{\sqrt{3} + u} \, du + \int \frac{1}{\sqrt{3} - u} \, du \right) \\[9pt] &= -\frac{\sqrt{3}}{2} \left( \log | \sqrt{3} + u | - \log | \sqrt{3} - u | \right). \end{align*}](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-a355e1d7399e757700649dbbea82db26_l3.png "Rendered by QuickLaTeX.com")
Putting these integrals back into our original expression we have
![\begin{align*} \int \frac{\sqrt{3-x^2}}{x} \, dx &= -\int \frac{x}{\sqrt{3-x^2}} \, dx + 3 \int \frac{1}{x \sqrt{3-x^2}} \, dx \\[10pt] &= \sqrt{3-x^2} - \frac{\sqrt{3}}{2} \left( \log | \sqrt{3} + u| - \log | \sqrt{3} - u| \right) \\[10pt] &= \sqrt{3-x^2} - \frac{\sqrt{3}}{2} \log \left| \frac{\sqrt{3} + \sqrt{3-x^2}}{\sqrt{3} - \sqrt{3-x^2}} \right| + C \\[10pt] &= \sqrt{3-x^2} - \frac{\sqrt{3}}{2} \log \left| \frac{(\sqrt{3} + \sqrt{3-x^2})^2}{x^2} \right| + C \\[10pt] &= \sqrt{3-x^2} - \sqrt{3} \log \left( \frac{\sqrt{3} + \sqrt{3-x^2}}{x} \right) + C. \end{align*}](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-b470ba2ecba9a998d58ab6a6898ff5ed_l3.png "Rendered by QuickLaTeX.com")
Using substitution of x=sqrt(3)sin(u) and dx=sqrt(3)cos(u) we can obtain Sqrt[3]Integrate[(Csc[u]-Sin[u]),u] , then with u=ArcSin[x/Sqrt[3]] to have x back, the same solution in achived.