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Question Number 186352 by normans last updated on 03/Feb/23

           ∫_1 ^( 2)   ((tan^(−1)  (x) + 2)/x^2 )  dx

12tan1(x)+2x2dx

Answered by MJS_new last updated on 04/Feb/23

∫((2+arctan x)/x^2 )dx=2∫(dx/x^2 )+∫((arctan x)/x^2 )dx  2∫(dx/x^2 )=−(2/x)+C_1   ∫((arctan x)/x^2 )dx=−((arctan x)/x)+∫(dx/(x(x^2 +1)))  ∫(dx/(x(x^2 +1)))=∫((1/x)−(x/(x^2 +1)))dx=(1/2)ln (x^2 /(x^2 +1)) +C_2   ⇒  answer is 1−((ln 5 −3ln 2)/2)+(1/2)arctan (1/2)

2+arctanxx2dx=2dxx2+arctanxx2dx2dxx2=2x+C1arctanxx2dx=arctanxx+dxx(x2+1)dxx(x2+1)=(1xxx2+1)dx=12lnx2x2+1+C2answeris1ln53ln22+12arctan12

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