# How do you integrate 1/(x^2 - 3)?

Sep 24, 2016

$= \frac{1}{2 \sqrt{3}} \ln \left\mid \frac{x - \sqrt{3}}{x + \sqrt{3}} \right\mid + C$

#### Explanation:

Decompose $\frac{1}{{x}^{2} - 3}$ using partial fractions:

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

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

$\frac{0 x + 1}{{x}^{2} - 3} = \frac{x \left(A + B\right) + \left(- \sqrt{3} A + \sqrt{3} B\right)}{{x}^{2} - 3}$

Thus:

$\left\{\begin{matrix}A + B = 0 \\ - \sqrt{3} A + \sqrt{3} B = 1\end{matrix}\right.$

Multiplying the first equation by $\sqrt{3}$:

$\left\{\begin{matrix}\sqrt{3} A + \sqrt{3} B = 0 \\ - \sqrt{3} A + \sqrt{3} B = 1\end{matrix}\right.$

Adding them gives $2 \sqrt{3} B = 1$ so $B = \frac{1}{2 \sqrt{3}}$.

Since $A + B = 0$, we see that $A = - \frac{1}{2 \sqrt{3}}$.

Thus:

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

$= - \frac{1}{2 \sqrt{3}} \ln \left\mid x + \sqrt{3} \right\mid + \frac{1}{2 \sqrt{3}} \ln \left\mid x - \sqrt{3} \right\mid$

$= \frac{\ln \left\mid x - \sqrt{3} \right\mid - \ln \left\mid x + \sqrt{3} \right\mid}{2 \sqrt{3}}$

$= \frac{1}{2 \sqrt{3}} \ln \left\mid \frac{x - \sqrt{3}}{x + \sqrt{3}} \right\mid + C$

Oct 2, 2016

$\frac{1}{2 \sqrt{3}} \ln \left\mid \frac{x - \sqrt{3}}{x + \sqrt{3}} \right\mid + C$

#### Explanation:

$I = \int \frac{\mathrm{dx}}{{x}^{2} - 3}$

Let $x = \sqrt{3} \sec \theta$, implying that $\mathrm{dx} = \sqrt{3} \sec \theta \tan \theta d \theta$:

$I = \int \frac{\sqrt{3} \sec \theta \tan \theta d \theta}{3 {\sec}^{2} \theta - 3} = \frac{\sqrt{3}}{3} \int \frac{\sec \theta \tan \theta d \theta}{{\sec}^{2} \theta - 1}$

Note that ${\sec}^{2} \theta - 1 = {\tan}^{2} \theta$:

$I = \frac{1}{\sqrt{3}} \int \frac{\sec \theta \tan \theta d \theta}{\tan} ^ 2 \theta = \frac{1}{\sqrt{3}} \int \frac{\sec \theta d \theta}{\tan} \theta = \frac{1}{\sqrt{3}} \int \frac{1}{\cos} \theta \left(\cos \frac{\theta}{\sin} \theta\right) d \theta = \frac{1}{\sqrt{3}} \int \csc \theta d \theta$

This is a common integral:

$I = - \frac{1}{\sqrt{3}} \ln \left\mid \csc \theta + \cot \theta \right\mid$

Note that $\sec \theta = \frac{x}{\sqrt{3}}$, thus we have a right triangle where $x$ is the hypotenuse, $\sqrt{3}$ is the side adjacent to $\theta$, and the opposite side is $\sqrt{{x}^{2} - 3}$.

Thus:$\text{ "csctheta=x/sqrt(x^2-3)" }$and$\text{ } \cot \theta = \frac{\sqrt{3}}{\sqrt{{x}^{2} - 3}}$

$I = - \frac{1}{\sqrt{3}} \ln \left\mid \frac{x}{\sqrt{{x}^{2} - 3}} + \frac{\sqrt{3}}{\sqrt{{x}^{2} - 3}} \right\mid$

$I = - \frac{1}{\sqrt{3}} \ln \left\mid \frac{x + \sqrt{3}}{\sqrt{{x}^{2} - 3}} \right\mid$

(Technical note. This almost could be a final answer, but there is one problem, which is that $\sqrt{{x}^{2} - 3}$ in the answer restricts the domain, in that we see that the domain excludes $- \sqrt{3} < x < \sqrt{3}$. This is not the case in the original function, but the trigonometry introduced this issue. To remedy this, add absolute value bars.)

$I = - \frac{1}{\sqrt{3}} \ln \left\mid \frac{x + \sqrt{3}}{\sqrt{\left\mid {x}^{2} - 3 \right\mid}} \right\mid$

This is a proper final answer. However, we can simplify it rather sneakily:

I=-1/sqrt3lnabs(sqrt((x+sqrt3)^2/(abs(x^2-3)))

The square root is a $\frac{1}{2}$ power, which can be brought from the logarithm rule: log(A^B)=Blog(A)#.

$I = - \frac{1}{2 \sqrt{3}} \ln \left\mid {\left(x + \sqrt{3}\right)}^{2} / \left({x}^{2} - 3\right) \right\mid$

$I = - \frac{1}{2 \sqrt{3}} \ln \left\mid {\left(x + \sqrt{3}\right)}^{2} / \left(\left(x + \sqrt{3}\right) \left(x - \sqrt{3}\right)\right) \right\mid$

$I = - \frac{1}{2 \sqrt{3}} \ln \left\mid \frac{x + \sqrt{3}}{x - \sqrt{3}} \right\mid$

We can also use the previous rule in reverse: $B \log \left(A\right) = \log \left({A}^{B}\right)$, to bring the $- 1$ power in and flip the fraction:

$I = \frac{1}{2 \sqrt{3}} \ln \left\mid \frac{x - \sqrt{3}}{x + \sqrt{3}} \right\mid + C$