Integrate $\int\frac{1}{x^2-3}dx$

Question

Integrate $\int\frac{1}{x^2-3}dx$

I'm trying to come into a form in order to apply this formula :

$\int\frac{1}{x^2-1}dx=\frac{1}{2}\cdot\ln|\frac{x-1}{x+1}|+C$

Need a bit help on this problem.

Thank you in advance :)

Answer 1

Probably the most straightforward way to proceed is to note that $$\frac{1}{x^2-3}=\frac{1}{\left(x+\sqrt3\right)\left(x-\sqrt3\right)}.$$ Next, find constants $A$ and $B$ such that $$\frac{1}{x^2-3}=\frac{A}{x+\sqrt3}+\frac{B}{x-\sqrt3}.$$ At that point, the individual antiderivatives should be clear, and you'll be able to condense into a single antiderivative using logarithm rules.

However, if you're bound and determined to use that formula, note that $$x^2-3=3\left(\frac{x^3}3-1\right)=3\left(\left(\frac{x}{\sqrt3}\right)^2-1\right),$$ so letting $u=\frac{x}{\sqrt3},$ we have $x=u\sqrt 3$ as a fine substitution to make.

Answer 2

Multiply and divide by 2√3 , then write 2√3 in the form of (x+√3)-(x-√3) , and split the denominator into (x+√3)(x-√3) , split the terms and now it's quite straightforward.