Find the integral?

Question

Find the integral of 4/(x^2+4) dx. 

Answer 1

The integrand  can be transformed to 1/(u²+1) and the integral of this tan^-1u..

Jim

Answer 2

Find the integral of 4/(x^2+4) dx

29963 (deleted): Find the area of the region bounded by the curve of
y=4/(x^2+4), the x-axis, and the vertical lines
x=-2 and x=2. Answer: pi

| Sun from Los Angeles, CA

A = integral{-2,2}[4/(x^2+2^2) dx]

arctan rule:
integral[du/(u^2+a^2)]=(1/a)arctan(u/a)+C
proof: http://www.matharticles.com/ma_clc7e.html > menu 5.9

for A, u=x,a=2:

A = 4 [1/2 arctan(x/2)]{-2,2}
= 2 [pi/4 - -pi/4]
= pi

Answer 3

Hi Sun

 

use x = 2 tan θ

 

1/(4 + 4tan²θ)dx

 

1/(4(1 + tan²θ)dx

 

so...dx/dθ = 2 tan θ

 

therefore, dx/dθ = 2 sec²θ

 

dx = 2 sec²θ dθ

 

so...1/(4 sec²θ) * 2 sec²θ dθ/1

 

1/2dθ = 1/2θ + c

 

x = 2 tan θ

 

x/2 = tan θ

 

tan^-1 (x/2) = θ

 

so... the integral of 1/(4 + x²) dx = 1/2 tan^-1 (x/2) + c