What is the improper integrals 1/(x^2+2x+2) dx from 0 to +infinity ? .

1 Answer
James
Jun 1, 2018

#color(blue)[int_0^(+oo)1/(x^2+2x+2)*dx=lim_(brarr+oo)int_0^b1/(x^2+2x+2)*dx=3pi/4]#

Explanation:

show below:

#int_0^(+oo)1/(x^2+2x+2)*dx=lim_(brarr+oo)int_0^b1/(x^2+2x+2)*dx#

#lim_(brarr+oo)int_0^b1/(x^2+2x+1+1)*dx=#

#lim_(brarr+oo)int_0^b1/((x+1)^2+1)*dx=lim_(brarr+oo)[arctan(x+1)]_0^b=#

#lim_(brarr+oo)[arctan(a+1)+arctan(1)]=[arctan(+oo)+pi/4]#

#pi/2+pi/4=3pi/4#

Related questions
Impact of this question
4304 views around the world
You can reuse this answer
Creative Commons License