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QUESTION Find $\ds\int\frac{dx}{2x^2+4x+3}$.


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ANSWER

$\ds\int\frac{dx}{2x^2+4x+3}=\int\frac{dx}{2(x^2+2x)+3}=
\int\frac{dx}{2(x+1)^2+1}\\ u =\sqrt{2}(x+1),\ du =\sqrt{2}dx.$

Therefore $\ds I =\int\frac{\frac{1}{\sqrt{2}}du}{u^2+1}
=\frac{1}{\sqrt{2}}\tan^{-1}u+c
=\frac{1}{\sqrt{2}}\tan^{-1}\{\sqrt{2}(x+1)\}+c$






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