How to solve $\int \frac{\mathrm dx}{3x^2-7}$?

Question

Do not know why but the integral drovw mw crazy $$\int \frac{\mathrm dx}{3x^2-7}$$

I tried $$\int(3x^2-7)^{-1}\,\mathrm dx$$ and then apply Chebyshev’s criterion, but it did not work :/

Answer 1

you can use this substitution: $x=\dfrac{\sqrt{7}}{\sqrt{3}}\sin u$ so $\mathrm dx= \dfrac{\sqrt{7}}{\sqrt{3}}\cos u$ and we have:

$$\int \dfrac{\frac{\sqrt{7}}{\sqrt{3}}\cos u}{7\cos^{2}u}\,\mathrm du=\frac{\sqrt{7}}{7\sqrt{3}}\int \sec u\,\mathrm du$$

which is easy to find.

Answer 2

Hint: Just express $\dfrac1{3x^2-7}$ as$$\frac13\left(\frac a{x-\sqrt{\frac73}}+\frac b{x+\sqrt{\frac73}}\right).$$

Answer 3

Hint: Use partial fractions

$$\int\dfrac{dx}{(\sqrt{3}x-\sqrt{7})(\sqrt{3}x+\sqrt{7})}=a\int\dfrac{dx}{\sqrt{3}x-\sqrt{7}}+b\int\dfrac{dx}{\sqrt{3}x+\sqrt{7}}$$

In which the coefficients can be directly determined by Euler's trick:

$$a=\left.\dfrac{1}{\sqrt{3}x+\sqrt{7}}\right|_{x=\dfrac{\sqrt{7}}{\sqrt{3}}}$$ $$b=\left.\dfrac{1}{\sqrt{3}x-\sqrt{7}}\right|_{x=-\dfrac{\sqrt{7}}{\sqrt{3}}}.$$