# #int_0^oo arctanx/(x(x^2+a^2))dx# ?

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We want to evaluate

#I=int_0^ooarctan(x)/(x(x^2+a^2))dx#

This can be solved by **differentiation under the integral sign**

Consider

#I(b)=int_0^ooarctan(bx)/(x(x^2+a^2))dx#

Notice

Differentiate both sides **w.r.t. b**

#I'(b)=int_0^oo1/((x^2+a^2)(x^2b^2+1))dx#

By partial fractions

#I'(b)=1/(1-a^2b^2)int_0^oo1/(x^2+a^2)-b^2/(x^2b^2+1)dx#

We recognize these integrals as

#I'(b)=1/(1-a^2b^2)[arctan(x/a)/a-b arctan(bx)]_0^oo#

Assuming

(We could just as well assumed

#I'(b)=1/(1-a^2b^2)((pi/2)/a-bpi/2)#

By some algebraic manipulation

#I'(b)=pi/2*1/(1-a^2b^2)((1-ab)/a)#

#color(white)(I'(b))=pi/(2a)(1-ab)/(1-a^2b^2)#

#color(white)(I'(b))=pi/(2a)1/(1+ab)#

Integrate both sides **w.r.t b**

#I(b)=pi/(2a)int1/(1+ab)db#

#color(white)(I(b))=pi/(2a^2)ln(1+ab)+C#

Evaluate the constant of integration (Remember

#0=pi/(2a^2)ln(1+a*0)+C=>C=0#

Thus

#I(b)=pi/(2a^2)ln(1+ab)#

Respecting negative values of

#I(b)=pi/(2a^2)ln(1+abs(a)b)#

In your case

#I(1)=I=pi/(2a^2)ln(1+abs(a))#

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