Complete the Square under the square root
\(\begin{aligned}{l}\int {\frac{1}{{\sqrt {9{x^2} + 6x - 8} }}dx} &= \int {\frac{1}{{\sqrt {{{(3x)}^2} + 2.(3x).1 + 1 - 1 - 8} }}} \\ &= \int {\frac{1}{{\sqrt {{{(3x + 1)}^2} - 9} }}dx} \end{aligned}\)
Substitute u = 3x+1
Let u = 3x+1
du = 3dx
\(\int {\frac{1}{{\sqrt {{{(3x + 1)}^2} - 9} }}dx = \frac{1}{3}\int {\frac{{du}}{{\sqrt {{u^2} - 9} }}} } \)
\(\begin{aligned}{l}\\\int {\frac{{dx}}{{\sqrt {{x^2} - {a^2}} }} = \ln \left| {x + \sqrt {{x^2} + {a^2}} } \right|} + c\end{aligned}\)
∴ \(\frac{1}{3}\int {\frac{{du}}{{\sqrt {{u^2} - 9} }} = \frac{1}{3}\ln \left| {x + \sqrt {{x^2} + {a^2}} } \right|} + c\)
Convert the value of the integral in terms of x
\(\begin{aligned}{l}\frac{1}{3}\ln \left| {3x + 1 + \sqrt {{{(3x + 1)}^2} - 9} } \right| + c\\\frac{1}{3}\ln \left| {3(x + \frac{1}{3}) + \sqrt {9{{(x + \frac{1}{3})}^2} - 1} } \right| + c\\\frac{1}{3}\ln \left| {3\left\{ {x + \frac{1}{3}} \right. + \left. {\sqrt {{{(x + \frac{1}{3})}^2} - 1} } \right\}} \right| + c\\\frac{1}{3}\ln \left| 3 \right| + \frac{1}{3}\ln \left| {x + \frac{1}{3} + \sqrt {{{(x + \frac{1}{3})}^2} - 1} } \right| + c\\\frac{1}{3}\ln \left| {x + \frac{1}{3} + \sqrt {{{(x + \frac{1}{3})}^2} - 1} } \right| + c'\\\end{aligned}\)
Final Answer:
Hence, \(\int {\frac{1}{{\sqrt {9{x^2} + 6x - 8} }}dx = \frac{1}{3}\ln \left| {x + \frac{1}{3} + \sqrt {{{(x + \frac{1}{3})}^2} - 1} } \right|} + c'\)
