For a real variable $x$, and $x>0$, integral expression for the modified Bessel function of second kind of order $\nu$ is $$K_\nu(x)=\int\limits_{0}^{\infty}e^{-x\cosh t}\cosh(\nu t)dt.$$ Let us consider the case $\nu=1$ such that $$K_1(x)=\int\limits_{0}^{\infty}e^{-x\cosh t}\cosh(t)dt.$$ Next, we make a change of variable: $$y=x\cosh(t)\Rightarrow dy=x\sinh(t) dt$$ so that the expression for $K_1(x)$ reduces to $$K_1(x)=x^{-1}\int\limits_{0}^{\infty}ye^{-y}(y^2-x^2)^{-1/2}dy.$$
But this does not agree with the expression in Eq.($11.69$) given here. In this expression $z$ is a real variable with $z\geq 0$, and is same as $x$ in my notation. What is the error I have made?
The lower limit of your (last) integral must be $x$. To get $(11.69)$ from it, just integrate by parts: $$xK_1(x)=\left.e^{-y}\sqrt{y^2-x^2}\right|_{y=x}^{y=\infty}-\int_x^{\infty}(-e^{-y})\sqrt{y^2-x^2}\,dy=\ldots$$