שינויים

<math>\int\frac{t+\frac{3}{42}}{t^2+\frac{3}{4}}\mathrm{d}t=\frac{1}{2}\int\frac{2t+\frac{3}{2}}{t^2+\frac{3}{4}}\mathrm{d}t</math>

<math>=\frac{1}{2}\int\frac{2t}{t^2+\frac{3}{4}}\mathrm{d}t+\int\frac{\frac{3}{42}}{t^2+\frac{3}{4}}\mathrm{d}t</math>

<math>=\frac{1}{2}\ln|t^2+\frac{3}{4}|+\frac{3}{42}\frac{1}{\sqrt{\frac{3}{4}}}\arctan\frac{t}{\sqrt{\frac{3}{4}}}+c</math>

<math>=\frac{1}{2}\ln|(x+\frac{1}{2})^2+\frac{3}{4}|+2\sqrt{\frac{3}{4}}\arctan\frac{x+\frac{1}{2}}{\sqrt{\frac{3}{4}}}+c</math>

ולכן <math>-\frac{2}{3}\int\frac{x+2}{x^2+x+1}\mathrm{d}x=-\frac{2}{3}(\frac{1}{2}\ln|(x+\frac{1}{2})^2+\frac{3}{4}|+2\sqrt{\frac{3}{4}}\arctan\frac{x+\frac{1}{2}}{\sqrt{\frac{3}{4}}})+c</math>

<math>x+\frac{2}{3}\ln|x-1|-\frac{2}{3}(\frac{1}{2}\ln|(x+\frac{1}{2})^2+\frac{3}{4}|+2\sqrt{\frac{3}{4}}\arctan\frac{x+\frac{1}{2}}{\sqrt{\frac{3}{4}}})+c</math>