Question

$$\displaystyle\int{ \frac{ 1 }{ \sqrt{ 16 { x }^{ 2 } -25 } } }d x$$

Answer

(d*x)/sqrt((4*x+5)*(4*x-5))

Solution


Rewrite \(16{x}^{2}-25\) in the form \({a}^{2}-{b}^{2}\), where \(a=4x\) and \(b=5\).
\[\int 1 \, \frac{d}{\sqrt{{(4x)}^{2}-{5}^{2}}}dx\]
Use Difference of Squares: \({a}^{2}-{b}^{2}=(a+b)(a-b)\).
\[\int 1 \, \frac{d}{\sqrt{(4x+5)(4x-5)}}dx\]
Use this rule: \(\int a \, dx=ax+C\).
\[\frac{dx}{\sqrt{(4x+5)(4x-5)}}\]