Question

how does the integral of dx/(xsqrt4x^2-1) become du/(usqrtu^2-1) when u=2x? specifically the 4x^2 becoming u^2

Answer 1

Bring the 4 into the square with the x,
\[\Large\rm \int\limits\limits \frac{dx}{x \sqrt{4x^2-1}}=\int\limits\limits \frac{dx}{x \sqrt{(2x)^2-1}}\]

Answer 2

Then we need to use these three pieces for our substitution:
\[\Large\rm u=2x, \qquad\qquad x=\frac{1}{2}u\qquad\qquad du=2dx\]

Answer 3

Would I should have written that third piece as: \(\Large\rm \dfrac{1}{2}du=dx\)
Plugging them in gives us:
\[\Large\rm \int\limits \frac{\frac{1}{2}du}{\frac{1}{2}u \sqrt{(u)^2-1}}\]

Answer 4

Woops* not would.. blah

Answer 5

thank you again!