Question

what os the formula of 1/x square-a square dx​

Answer 1

Step-by-step explanation:

I will suppose you aske (1/x)^2 - (a)^2

(1/x)^2 - (a)^2. ( Using a^2 - b^2 = (a+b)(a-b))

= (1/x + a)(1/x - a )

Answer 2

Answer:

Integral of Some Particular Functions

Integral of Some Particular FunctionsLook at the following integration formulas

1. Integral of Some Particular FunctionsLook at the following integration formulas∫ dx / (x2 – a2) = 1/2a log |(x – a) / (x + a)| + C
2. Integral of Some Particular FunctionsLook at the following integration formulas∫ dx / (x2 – a2) = 1/2a log |(x – a) / (x + a)| + C∫ dx / (a2 – x2) = 1/2a log |(a + x) / (a – x)| + C
3. Integral of Some Particular FunctionsLook at the following integration formulas∫ dx / (x2 – a2) = 1/2a log |(x – a) / (x + a)| + C∫ dx / (a2 – x2) = 1/2a log |(a + x) / (a – x)| + C∫ dx / (x2 + a2) = 1/a tan–1 (x/a) + C
4. Integral of Some Particular FunctionsLook at the following integration formulas∫ dx / (x2 – a2) = 1/2a log |(x – a) / (x + a)| + C∫ dx / (a2 – x2) = 1/2a log |(a + x) / (a – x)| + C∫ dx / (x2 + a2) = 1/a tan–1 (x/a) + C∫ dx / √ (x2 – a2) = log |x + √ (x2 – a2)| + C
5. Integral of Some Particular FunctionsLook at the following integration formulas∫ dx / (x2 – a2) = 1/2a log |(x – a) / (x + a)| + C∫ dx / (a2 – x2) = 1/2a log |(a + x) / (a – x)| + C∫ dx / (x2 + a2) = 1/a tan–1 (x/a) + C∫ dx / √ (x2 – a2) = log |x + √ (x2 – a2)| + C∫ dx / √ (a2 – x2) = sin–1 (x/a) + C
6. Integral of Some Particular FunctionsLook at the following integration formulas∫ dx / (x2 – a2) = 1/2a log |(x – a) / (x + a)| + C∫ dx / (a2 – x2) = 1/2a log |(a + x) / (a – x)| + C∫ dx / (x2 + a2) = 1/a tan–1 (x/a) + C∫ dx / √ (x2 – a2) = log |x + √ (x2 – a2)| + C∫ dx / √ (a2 – x2) = sin–1 (x/a) + C∫ dx / √ (x2 + a2) = log |x + √ (x2 + a2)| + C