อ้างอิง:
ข้อความเดิมเขียนโดยคุณ JamesCoe#18
ข้อ $\int ln(x^2+x+1)dx$
ไม่เฉลยให้ดูหน่อยหรอคับคุณV.Rattanapon
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\[
\int {\ln \left( {x^2 + x + 1} \right)dx = x\ln } \left( {x^2 + x + 1} \right) - \int {\frac{{2x^2 + x}}{{x^2 + x + 1}}dx}
\]
\[
= x\ln \left( {x^2 + x + 1} \right) - \int {\left( {2 - \frac{1}{2}\left( {\frac{{2x + 1}}{{x^2 + x + 1}}} \right) - \frac{3}{2}\left( {\frac{1}{{x^2 + x + 1}}} \right)} \right)} dx
\]
\[
= x\ln \left( {x^2 + x + 1} \right) - \left( {2\int {dx - \frac{1}{2}\int {\left( {\frac{1}{{x^2 + x + 1}}} \right)d\left( {x^2 + x + 1} \right) - \frac{3}{2}\int {\frac{{d\left( {x + \frac{1}{2}} \right)}}{{\left( {x + \frac{1}{2}} \right)^2 + \left( {\frac{{\sqrt 3 }}{2}} \right)^2 }}dx} } } } \right)
\]
\[
= x\ln \left( {x^2 + x + 1} \right) - 2x + \frac{1}{2}\ln \left| {x^2 + x + 1} \right| + \sqrt 3 \arctan \left( {\frac{{2x + 1}}{{\sqrt 3 }}} \right) + c
\]