Question

integral de 1/(4-9x²) . dx

Answer 1

$\int\limits {\dfrac{1}{4-9x^2}} \, dx =\\\\\\\int\limits {\dfrac{1}{4\left(1-\frac{9x^2}{4}\right)}} \, dx \\\\\\\cdot u=\dfrac{3x}{2}\to du=\dfrac32\ dx\to dx=\dfrac23\ du\\\\\\\int\limits {\dfrac{1}{4\left(1-u^2\right)}} \,\dfrac23\ du=\\\\\\\dfrac{1}{6}\int\limits {\dfrac{1}{\left(1-u^2\right)}}\ du=\\\\\\\dfrac{1}{6}\int\limits {\dfrac{1}{(1+u)(1-u)}}\ du=\\\\\\\dfrac{1}{6}\int\limits {\dfrac{A}{1+u}+\dfrac{B}{1-u}}\ du$

$\cdot A-Au+B+Bu=1\\\\\left \{ {{A+B=1} \atop {B-A=0}} \right. \\\\B=\dfrac12,\ A=-\dfrac12\\\\\\\dfrac16\int\limits {-\dfrac12\dfrac{1}{1+u}+\dfrac12\dfrac{1}{1-u}} \, du =\\\\\\\dfrac{1}{12}\int\limits {\dfrac{1}{1+u}+\dfrac{1}{1-u}} \, dx =\\\\\\\dfrac{1}{12}(\ln(1+u)-\ln(1-u))+C=\\\\\\\boxed{\dfrac{1}{12}\left(\ln\left(1+\dfrac{3x}{2}\right)-\ln\left(1-\dfrac{3x}{2}\right)\right)+C}$