Question Number 32258 by Cheyboy last updated on 22/Mar/18
$${find} \\ $$$$\int\:\frac{\mathrm{1}}{\mathrm{2}−{x}^{\mathrm{2}} }\:{dx} \\ $$
Answered by mrW2 last updated on 22/Mar/18
![∫ (1/(((√2))^2 −x^2 )) dx ∫ (1/([(√2)+x][(√2)−x])) dx −(1/(2(√2)))[∫ (1/(x−(√2))) dx−∫(1/(x+(√2))) dx] −(1/(2(√2)))ln ∣((x−(√2))/(x+(√2)))∣+C](https://www.tinkutara.com/question/Q32259.png)
$$\int\:\frac{\mathrm{1}}{\left(\sqrt{\mathrm{2}}\right)^{\mathrm{2}} −{x}^{\mathrm{2}} }\:{dx} \\ $$$$\int\:\frac{\mathrm{1}}{\left[\sqrt{\mathrm{2}}+{x}\right]\left[\sqrt{\mathrm{2}}−{x}\right]}\:{dx} \\ $$$$−\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}\left[\int\:\frac{\mathrm{1}}{{x}−\sqrt{\mathrm{2}}}\:{dx}−\int\frac{\mathrm{1}}{{x}+\sqrt{\mathrm{2}}}\:{dx}\right] \\ $$$$−\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}\mathrm{ln}\:\mid\frac{{x}−\sqrt{\mathrm{2}}}{{x}+\sqrt{\mathrm{2}}}\mid+{C} \\ $$
Commented by Cheyboy last updated on 23/Mar/18
$${Thank}\:{you}\:{sir}\:{Godbless}\:{you} \\ $$
Answered by $@ty@m last updated on 22/Mar/18
