Calculus: Early Transcendentals, Chapter 7, 7.1, Section 7.1, Problem 30

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)