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We have the integral $\displaystyle\int \dfrac{1}{x^2+4x+3} dx$. I figured out that with a substitution you can rewrite this as $ - \displaystyle \int \dfrac{1}{1-u^2} du$, which is $-\tanh^{-1}(u)$. However, I wonder if there's another way to do this integral without having to use the (in my opinion) obscure inverse hyperbolic trig functions.

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  • $\begingroup$ Inverse hyperbolic tanh = $$\frac12 \log{\left ( \frac{1-u}{1+u} \right )}$$ $\endgroup$ – Ron Gordon Nov 9 '15 at 20:27
  • $\begingroup$ @RonGordon I wasn't familiar with that identity $\endgroup$ – Miletti Nov 9 '15 at 20:38
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You can apply parital fractions to this problem and then simply integrate fractions by logarithms.

$$\frac{1}{x^2+4x+3}=\frac{1}{2(x+1)}-\frac{1}{2(x+3)}$$

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$$\frac{1}{x^2+4x+3}=\frac{1}{(x+1)(x+3)}=\frac{1}{2}(\frac{1}{x+1}-\frac{1}{x+3})$$ and then use the $$\int \frac{1}{x+a}dx=\log|(x+a)|+C$$

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