Calculate the integral: (dx)/(x^(2)+25) between - negative and + infinity?

Question

Calculate the integral: (dx)/(x^(2)+25) between - negative and + infinity

If they converge, it is said I should appeal to the dominance of one function over the other , or also by applying l"Hopital's rule.

Any help is greatly appreciated!

Answer 1

First you know don't you that you use the substitution x = 5 tan u, so dx/du = 5 sec²u = 5 (1 + tan²u) = 5 (1 + x²/25).



So the integral becomes du.5(1 + x²/25) / (25 + x²) = 5 du, so is 5u + c = 5 arctan (x/5) + c.



Now when x is infinity, this is 5 (π/2) and when minus infinity, 5 (-π/2), so definite integral is 5π.

Answer 2

∫ dx/(x^2+25)

= (1/25) ∫ dx/(x^2/25 +1)

= (1/25) ∫ dx/( (x/5)^2 +1)



Let u= x/5

du = dx/5

dx= 5 du

= (1/25) ∫ dx/( (x/5)^2 +1) = (1/5) ∫ du/(u^2+1) = (1/5) tan^-1(u)



= (1/5) tan^-1( x/5)



Let F(x) = (1/5) tan^-1( x/5)



substitute the upper limit ∞

F(∞) = lim x-->∞ (1/5) tan^-1( x/5) = (1/5) (pi/2)



substitute the lower limit -∞

F(-∞) = lim x-->-∞ (1/5) tan^-1( x/5) = (1/5) (-pi/2)



subtract:

F(∞)-F(-∞) = pi/10 - (-pi/10) = pi/5

Answer 3

pi/5

Step by step answer

http://www.mathskey.com/question2answer/21147/calculate-the-integral-dx-x-2-25