Answer
$(\dfrac{1}{3}) \ln |\dfrac{\sqrt {x+1}-1}{\sqrt {x+1}+1}|+c$
Work Step by Step
Consider the integral $\int \dfrac{dx}{x(3 \sqrt {x+1})}$
Let us take the help of the substitution method.
$a= \sqrt {x+1} \implies da=\dfrac{dx}{2 \sqrt {x+1}}; dx=(2a) da$
The given integral can be re-written as: $\int \dfrac{dx}{x(3 \sqrt {x+1})}= \dfrac{2}{3}\int \dfrac{da}{(a^2-1)}=\dfrac{1}{3}\int \dfrac{da}{(a-1)}-\dfrac{1}{3}\int \dfrac{da}{(a+1)}$
This implies that $\int [(2x-3)+ \dfrac{x}{x^2+2x-8}]dx=\int (2x-3) dx+(\dfrac{1}{3})\int \dfrac{dx}{x-2}+(\dfrac{2}{3}) \int \dfrac{dx}{x+4}$
We use the formula: $\int x^n dx=\int\dfrac{x^{n+1}}{n+1} dx$
or, $\dfrac{1}{3} \ln |a-1|-\dfrac{1}{3} \ln |a+1|+C=(\dfrac{1}{3}) \ln |\dfrac{\sqrt {x+1}-1}{\sqrt {x+1}+1}|+c$
