How do you find the integral of int 1/(xsqrt(4x^2-1)dx?

1 Answer
Jan 26, 2017

The answer is =arctan(sqrt(4x^2-1)) + C

Explanation:

Let's do this integral by substitution

Let u=sqrt(4x^2-1)

(du)/dx=1/(2sqrt(4x^2-1))*8x=(4x)/(sqrt(4x^2-1))

Also,

u^2=4x^2-1

4x^2=u^2+1

Therefore,

int(dx)/(xsqrt(4x^2-1))=int(cancelsqrt(4x^2-1)du)/(4x*x*cancelsqrt(4x^2-1))

=int(du)/(u^2+1)

Let u=tantheta

u^2+1=sec^2theta

du=sec^2theta d theta

Therefore,

int(du)/(u^2+1)=int(sec^2theta d theta)/(sec^2 theta)

intd theta=theta=arctanu

So,

int(dx)/(xsqrt(4x^2-1))=arctan(sqrt(4x^2-1)) + C