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int 2-> infinity (1/(x^2 +4)) dx
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int 1-> infinity (1/(x^2 +4)) dx
\[\int\limits_{}1/(x^{2}+4)dx=1/2 \int\limits_{}(1/2)/((x/2)^{2}+1)dx=ArcTan(x/2)/2\] is this what you're asking?
\[\int\limits_{1}^{\infty}1/(x2+4)dx \]
well that gives ArcTan(inf/2)/2 - ArcTan(1/2)/2=pi/4-ArcTan(1/2)
Since ArcTan(inf)=pi/2
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