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Let 2x=tanθ, 2dx=sec2θdθ, 1/(1+(2x)2)=1/(1+tan2θ)=1/sec2θ,

dx/(1+(2x)2)=½sec2θdθ/sec2θ=½dθ.

∫½dθ=θ/2+C=½tan-1(2x)+C where C is integration constant.

This can also be written ½arctan(2x)+C.

by Top Rated User (1.1m points)

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