Question

integration of dx/4-x^2

Answer 1

Step-by-step explanation:

I==integ.of dx/(4+x^2)

Let x=2.tanA. , dx= 2.sec^2A.dA

I=integ.of 2.sec^2A.dA/(4+4tan^2A).

I=integ.of 2.sec^2A.dA/4.(1+tan^2A)

I=integ.of 2.sec^2A.dA/4.sec^2A

I=integ.of (1/2).dA

I=(1/2).A+C

I=1/2.tan^-1(x/2). + C. Answer.

Answer 2

Given,

I = int dx/ 4 + x^2

I = int dx/ 2^2 + x^2

I = int dx / x^2 +2^2

I = (1/2) tan^-1 (x/2)

using formula

• int . dx/ x^2 +a^2 = (1/a) tan^-1 (x/a)

I = 0.5 tan^-1 (x/2)