Question

Как вычислить интеграл?

Answer 1

$\int\limits\frac{dx}{4x^2-1}=\int{\dfrac{4}{\left(4\,x-2\right)\,\left(4\,x+2\right)}}{\;\mathrm{d}x}\overset{4x=u}{=}\int{\dfrac{1}{\left(u-2\right)\,\left(u+2\right)}}{\;\mathrm{d}u}=$

$=-\frac{1}{4}\int \left ( \frac{1}{u+2}-\frac{1}{u-2} \right )\mathrm{d}u=\dfrac{\ln\left(\left|u-2\right|\right)}{4}-\dfrac{\ln\left(\left|u+2\right|\right)}{4}+C=\dfrac{\ln\left(\left|u-2\right|\right)}{4}-\dfrac{\ln\left(\left|u+2\right|\right)}{4}+C$

$\int\limits_{1}^{3}\frac{dx}{4x^2-1}=\dfrac{\ln\left(5\right)}{4}-\dfrac{\ln\left(7\right)}{4}-\left ( -\dfrac{\ln\left(3\right)}{4}\right )=-\dfrac{\ln\left(7\right)}{4}+\dfrac{\ln\left(5\right)}{4}+\dfrac{\ln\left(3\right)}{4}=\frac{1}{4}\ln \left ( \frac{15}{7} \right )$