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integrate ( (2x+1).(x^2+x+1)dx ) / ( (x^2+x+1)^2 + (x^2+x+1) ) @MIT 18.01 Single …
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Divide top and bottom by x^2+x+1 to give: \[(2x+1)/x ^{2}+x+2\] Use substitution \[u=x^2+x+2\]which means that \[du=(2x+1)dx\]Hence the integral becomes \[\int\limits_{}^{}(1/u)du\]which is \[\ln|u| + C = \ln|x^2+x+2| + C\]
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