How do you find the integral of #int (1/(1+x^2) ) dx# from negative infinity to 0?

Answer 1

#pi/2#

let #x=tantheta# #dx=sec^2theta d(theta)# #1+tan^2theta=sec^2theta# #if x=-oo theta=-pi/2# #if x=9 theta=0# #int_-oo^01/(1+x^2)dx=int_-(pi/2)^0sec^2theta/sec^2thetad(theta)# =#int_-(pi/2)^0d(theta)#=#[theta]_(-pi/2)^0=0-(-pi/2)=pi/2#
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Answer 2
This is related to #arctanx#...
#d/(dx)[arctanx] = 1/(1+x^2)#
#:. " " int_a^b 1/(1+x^2)dx = arctanx + C#
#int_(-oo)^(0) 1/(1+x^2) = arctan(0) - arctan(-oo)#
Due to #tanx# having a domain of #(pi/2, pi/2) pm pik# where #k in ZZ#, the inverse, #arctanx#, is a graph with a range of #(-pi/2,pi/2)#, and it looks kind of like a sideways #x^3#, crossing through #(0,0)#.

graph{arctanx [-20, 20, -3.12, 3.12]}

Therefore, as #x->-oo#, you get #-pi/2#, and so:
#= 0 - (-pi/2) = color(blue)(pi/2)#
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Answer 3

To find the integral of ( \int_{-\infty}^{0} \frac{1}{1+x^2} , dx ), you can use the following approach:

  1. Recognize that the integrand is a standard form of the arctangent function.
  2. Apply the property of the arctangent function: ( \int \frac{1}{1+x^2} , dx = \arctan(x) + C ).
  3. Evaluate the definite integral using the limits of integration: ( \arctan(0) - \lim_{a \to -\infty} \arctan(a) ).
  4. Since ( \arctan(0) = 0 ) and ( \lim_{a \to -\infty} \arctan(a) = -\frac{\pi}{2} ), the result is ( 0 - (-\frac{\pi}{2}) = \frac{\pi}{2} ).

Therefore, the integral of ( \int_{-\infty}^{0} \frac{1}{1+x^2} , dx ) equals ( \frac{\pi}{2} ).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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