How to compute $\int {\frac{1}{{4 - 9{x^2}}}dx} $?

Question

How can I evaluate the following integral

$$\int {\frac{1}{{4 - 9{x^2}}}dx} $$

Answer 1

Notice, use partial fractions as follows $$\int \frac{dx}{4-9x^2}=\int \frac{dx}{(2-3x)(2+3x)}$$ $$=\int \frac{1}{4}\left(\frac{1}{2-3x}+\frac{1}{2+3x}\right)\ dx$$ $$=\frac{1}{4}\int \left(\frac{1}{2-3x}+\frac{1}{2+3x}\right)\ dx$$

Answer 2

Hint:

The derivative of $\tanh^{-1}(x)$ is $$\frac 1{1-x^2}.$$ Then the integral will be of the form $$a\tanh^{-1}(bx).$$

Answer 3

Isn't it this term reminding arctanh derivative ? :-) for $\frac{1}{a+b x^2}$ -like integrands, first re-check $arctan(ax)$ and $arctanh(ax)$ derivatives, choose the best appropriate, then fit to your numbers.