Solve if the improper integral below diverges or converges. If it converges, solve for the value. La 1 e* dx (1 —еx)2 1

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Q: -+inf In(tan¯1x) 1+x2 1

A: ∫1∞ln(arctan(x)1+x2dxSubstiute u=arctan(x)→dudx=11+x2→dx=(x2+1)du=∫ln(u)duIntegrate by parts…