Issolvi |2x+1|-1/2ln(x^2+x+1)+1/sqrt{3}arctan2x/1

$$| 2 x + 1 | - \frac { 1 } { 2 } \ln ( x ^ { 2 } + x + 1 ) + \frac { 1 } { \sqrt { 3 } } \arctan \frac { 2 x } { 1 }$$

Fattur

$$\frac{-3\ln(x^{2}+x+1)+2\sqrt{3}\arctan(2x)+6|2x+1|}{6}$$

Passi tas-Soluzzjonijiet

Irrazzjonalizza d-denominatur tal-\frac{1}{\sqrt{3}} billi timmultiplika in-numeratur u d-denominatur mill-\sqrt{3}.

$$factor(|2x+1|-\frac{1}{2}\log_{e}\left(x^{2}+x+1\right)+\frac{\sqrt{3}}{\left(\sqrt{3}\right)^{2}}\arctan(\frac{2x}{1}))$$

Il-kwadrat ta' \sqrt{3} huwa 3.

$$factor(|2x+1|-\frac{1}{2}\log_{e}\left(x^{2}+x+1\right)+\frac{\sqrt{3}}{3}\arctan(\frac{2x}{1}))$$

Kwalunkwe ħaġa diviża b'wieħed tagħti riżultat tagħha stess.

$$factor(|2x+1|-\frac{1}{2}\log_{e}\left(x^{2}+x+1\right)+\frac{\sqrt{3}}{3}\arctan(2x))$$

Esprimi \frac{\sqrt{3}}{3}\arctan(2x) bħala frazzjoni waħda.

$$factor(|2x+1|-\frac{1}{2}\log_{e}\left(x^{2}+x+1\right)+\frac{\sqrt{3}\arctan(2x)}{3})$$

Iffattura 'l barra \frac{1}{6}.

$$\frac{6|2x+1|-3\log_{e}\left(x^{2}+x+1\right)+2\sqrt{3}\arctan(2x)}{6}$$

Evalwa

$$\frac{-3\ln(x^{2}+x+1)+2\sqrt{3}\arctan(2x)+6|2x+1|}{6}$$