Question

integrate 1/(x^2 Sqrt[4 x^2 + 1])

Answer 1

Let 2x=tan(y), dx=sec^2(y)dy/2.

√(4x^2+1)=√(tan^2(y)+1)=√sec^2(y)=sec(y).

The integrand becomes:

4sec^2(y)dy/2tan^2(y)sec(y)=2sec(y)dy/tan^2(y)=2cot(y)cosec(y)dy.

Differential of cosec(y)=differential of sin^-1(y)=

-sin^-2(y)cos(y)=-cot(y)cosec(y).

So ∫2cot(y)cosec(y)dy=-2cosec(y)+C=-√(4x^2+1)/x+C (where C is the constant of integration).

(This can also be written C-√(4+(1/x^2)). Differentiating we get -(1/2)(4+x^-2)^-(1/2)*(-2x^-3)=(1/(x^3√(4+x^-2)). In the denominator we take one x into the square root: 1/(x^2√(4x^2+1)). This is the original integrand, which confirms the integration solution.)

Answer 2

I dont think there is  a solition for this Integral . If the symbol was - not + and the integral is

 Integral = I

I { 1/(x^2*sgrt(4x^2-1))dx} =

x = secu/2 and dx = secutanudu/2

I secutanudu/(2(secu/2)^2*sgrt(4(secu/2)^2 -1)) du=

 2*I tanu/(secu*sgrt(sec^2u  - 1)) du=

2 I {sin/cos)/((sgrtsec^2u- 1)/cosu)) *du =

2 I (sinu/(sgrt(sec^2u - 1)) du= 2 I cos du = 2 sinu + c = 2 sin(arcsecu) + c = 2sin(arcsec(2x)) + c

sinu/(sgrt(sec^2-1)) = sinu/((sgrt(1/cos^2u -1))= sinu/((1-cos^2u)/cos^2u)=

sinucosu/sinu = cosu