Important Probleme integrale

∫ 1/(2+sinx) dx
∫(2x+1)/( x^2 -x+1) dx

Buna seara,

1. Daca \(tg\frac{x}{2}=t\), atunci \(sinx=\frac{2t}{1+t^{2}}\); \(x=2arctgt\Rightarrow dx=\frac{2}{t^{2}+1}\).

Atunci \(I=\int \frac{1}{2+sinx}dx=\int \frac{1}{2+\frac{2t}{1+t^{2}}}\cdot \frac{2}{1+t^{2}}dt=\int \frac{1}{t^{2}+t+1}dt\).

\(t^{2}+t+1=\left ( t+\frac{1}{2} \right )^{2}+\left (\frac{\sqrt{3}}{2} \right )^{2}\), deci
\(I=\frac{1}{\frac{\sqrt{3}}{2}}\cdot arctg\frac{t+\frac{1}{2}}{\frac{\sqrt{3}}{2}}+C=\frac{2arctg\frac{\left ( 2tg\frac{x}{2}+1 \right )}{\sqrt{3}}}{\sqrt{3}}+C\)

2. Aici incep rezolvarea, sa completati Dvs. mai departe. Postati si rezultatul, va rog.

\(\frac{2x+1}{x^{2}-x+1}=\frac{2x-1}{x^{2}-x+1}+\frac{2}{x^{^{2}}-x+1}\) \(=\frac{\left ( x^{2}-x+1 \right )'}{x^{2}-x+1}+\frac{2}{\left ( x-1 \right )^{2}+\left ( \frac{\sqrt{3}}{2} \right )^{2}}\).

Atunci \(\int \frac{2x+1}{x^{2}-x+1}dx=\int \frac{\left ( x^{2}-x+1 \right )'}{x^{2}-x+1}dx+\int \frac{2}{\left ( x-\frac{1}{2} \right )^{2}+\left ( \frac{\sqrt{3}}{2} \right )^{2}}dx\) = ...