intx/sqrt(3x²-6x-8)dx
=1/sqrt(3)intx/sqrt(x²-2x-8/3)dx
=1/sqrt(3)intx/sqrt((x-1)²-11/3)
Let x-1=sqrt(11/3)csc(theta)
dx=-sqrt(11/3)cot(theta)csc(theta)d theta
1/sqrt(3)intx/sqrt((x-1)²-11/3)=-1/sqrt(3)int((sqrt(11/3)csc(theta)+1)cot(theta)csc(theta))/sqrt(csc²(theta)-1)d theta
Because csc²(x)-1=cot²(x),
=-1/sqrt(3)(intsqrt(11/3)csc²(theta)d theta + intcsc(theta)d theta)
=sqrt11/3cot(theta)+1/sqrt3ln(|cot(theta)+csc(theta)|)
Finally, theta=csc^(-1)(sqrt(3/11)(x-1))
So:
intx/sqrt(3x²-6x-8)dx=sqrt11/3cot(csc^(-1)(sqrt(3/11)(x-1)))+1/sqrt3ln(|cot(csc^(-1)(sqrt(3/11)(x-1)))+sqrt(3/11)(x-1)|)+C, C in RR
\0/ here's our answer !