∫(x+1)/√(x^2+x+1)dx
∫x+1√x2+x+1dx
∫x+1√x2+x+1dx=∫x+1√(x+12)2+34dx=2√3∫x+1√(2x+1)23+1dx=2√3∫√3u−12+1√u2+1d√3u−12⋅(u=2x+1√3,x=√3u−12)=12∫√3u+1√u2+1du=√32∫u√u2+1du+12∫1√u2+1du=√x2+x+1+12arcsinh√33(2x+1)+C
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最后修改于2023年05月04日
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