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

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Answer 1 · 2017-01-26T09:49:02

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

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#