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\int\limits (8x)/(x^2+1) dx

User Steve Pettifer
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1 Answer

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\displaystyle \int~\cfrac{8x}{x^2+1}~dx~\hfill u=x^2+1\implies \cfrac{du}{dx}=2x\implies \cfrac{du}{2x}=dx \\\\\\ \displaystyle \int~\cfrac{8x}{u}\cdot \cfrac{du}{2x}\implies \int~\cfrac{4}{u}\cdot du\implies 4\int~\cfrac{1}{u}~du \\\\\\ 4\ln|u|+C\implies {\Large \begin{array}{llll} 4\ln|x^2+1|+C \end{array}}

User Smack
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