`int_1^sqrt(3)arctan(1/x)dx`

If f(x) and g(x) are differentiable functions, then

`intf(x)g'(x)dx=f(x)g(x)-intf'(x)g(x)dx`

If we write f(x)=u and g'(x)=v, then

`intuvdx=uintvdx-int(u'intvdx)dx`

Using the above method of integration by parts,

`intarctan(1/x)dx=arctan(1/x)*int1dx-int(d/dx(arctan(1/x)int1dx)dx`

`=arctan(1/x)*x-int(1/(1+(1/x)^2)*d/dx(1/x)int1dx)dx`

`=xarctan(1/x)-int(x^2/(x^2+1)*(-1x^-2)*x)dx`

`=xarctan(1/x)+intx/(x^2+1)dx`

Now let's evaluate `intx/(x^2+1)dx`

substitute `t=x^2+1,=>dt=2xdx`

`intx/(x^2+1)dx=intdt/(2t)`

`=1/2ln|t|`

substitute back `t=x^2+1`

`=1/2ln|x^2+1|`

`intarctan(1/x)dx=xarctan(1/x)+1/2ln|x^2+1|+C`

C is a constant

Now evaluate the definite integral,

`int_1^sqrt(3)arctan(1/x)dx=[xarctan(1/x)+1/2ln|x^2+1|]_1^sqrt(3)`

`=[sqrt(3)arctan(1/sqrt(3))+1/2ln(3+1)]-[1arctan(1/1)+1/2ln(1+1)]`

`=[sqrt(3)pi/6+1/2ln(4)]-[pi/4+1/2ln2]`

`=[sqrt(3)pi/6+1/2ln(2^2)]-[pi/4+1/2ln(2)]`

`=(sqrt(3)pi/6+ln(2)-pi/4-1/2ln(2))`

`=(sqrt(3)pi/6-pi/4+1/2ln(2))`

`=(2sqrt(3)-3)pi/12+1/2ln(2)`

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