Question

Answer 1

b: \(\frac{1}{x^2+x+1}=\frac{1}{\left(x+\frac12\right)^2+\frac34}\)

Đặt \(u=x+\frac12\)

=>du=dx

\(\int\frac{1}{x^2+x+1}\cdot\left(dx\right)=\int\frac{1}{u+\frac34}du=\ln\left|u+\frac34\right|=\ln\left|x+\frac12+\frac34\right|=\ln\left|x+\frac54\right|\)

=>\(\int_{-1}^1\!\frac{1}{x^2+x+1}\,\mathrm{d}x=\ln\left|1+\frac54\right|-\ln\left|-1+\frac54\right|=\ln\left|\frac94\right|-\ln\left|\frac14\right|=\ln\left|9\right|=\ln9\)