In Rudin's "Real & Complex Analysis", he states a special case (p.156 in my book) of Theorem 7.26: If $X:[\alpha,\beta] \rightarrow [a,b]$ is AC and monotonic, with $x(\alpha) = a$, $x(\beta) = b$, and (Lebesque) measurable $f\geq 0$, then $\int_{[a,b]} f = \int_{[\alpha,\beta]} f \circ X \; X'$ Since $f$ is integrable, we may split $f$ into positive and non-positive parts, apply the above to both separately, and then stitch them back together again. (Integrability ensures there is no 'funny' business with infinities.)
Another approach would be to take the family of parametrizations $X_{\epsilon}$ defined by $t \mapsto X(t) + \epsilon t$, and let $\epsilon \rightarrow 0$. When $\epsilon>0$, $X_{\epsilon}$ is strictly increasing. It is straightforward to show that the left hand side of the equality converges to the expected value. The right hand side requires more work.
$X$ can be used to define a measure that is AC with respect to the Lebesque measure. We have the convenient relationship $X_{\epsilon}'(t) = X'(t) + \epsilon$. Then $f$ can be approximated by a continuous function first with respect to the measure $X$ and then the Lebesque measure. Since the continuous approximations are uniformly continuous, one can show that the mapping $\epsilon \mapsto \int_{[\alpha,\beta]} f \circ X_{\epsilon} \; X_{\epsilon}'$ is continuous, from which the desired result follows.
Note that much of the work in Rudin is involved in showing measurability of the integrand on the right hand side. This is much less of an issue here, because the change of variables $X$ has a nice form.