Integration?

Question

How to integrate "dx/5+x^(2)" and "dx/(x)^2-(2)^2".Here two problems and both are indefinite integrals.Give answer with explaination.

Answer 1

(1)

Using the formula: ⌠1/ (1+x²) dx = tan^-1 + c

Given:
⌠1 / (5 + x²) dx
(1/5)⌠1 / (1/5)(5 + x²) dx
(1/5)⌠1 /(1 + (x²/5)) dx
(1/5)⌠1 /(1 + (x/25)²) dx
= (1/5)[(tan^-1 (x/25)] 1/(1/25) + c
= 5 [(tan^-1 (x/25)] + c

Going by the integration formula:

⌠1/ (a²-x²) = 1/2a ln |(a+x)/(a-x)| + c
-⌠1/ - (a²-x²) = 1/2a ln |(a+x)/(a-x)| + c
-⌠1/ x²-a² = - [1/2a ln |(a+x)/(a-x)| ] - c
Given:
⌠1/ (a²-x²) dx
-⌠1/ x²-2² = - [1/2(2) ln |(2+x)/(2-x)| ] - c
-⌠1/ x²-2² = - [1/4 ln |(2+x)/(2-x)| ] - c

Answer 2

intgrate dx/(x^2+5)

let x = 5^(1/2) tan t

dx = 5^1/2(sec t) ^2
x^2 +5 = 5 (sect)^2

so integral = int(5^.5) dt = 5^.5 t = t^.5 arctan(x/5^2)

Answer 3

x^2+5 = x^2 + (sqrt(5))^2

Its integral is atan(x/sqrt(5))/sqrt(5)

1/(x^2-2^2) = 1/(x-2)(x+2) = 1/4(1/(x-2) - 1/(x+2))

Its integral is

1/4 log((x-2)/(x+2))