Integral Table Result To prove the result
$$\int_{1}^{e}\frac{\ln(x)}{(\alpha +\ln(x))^{(\alpha+1)}}\,dx = \frac{e}{(\alpha+1)^\alpha} - \frac{1}{\alpha^\alpha}$$
I have used the substitution $t = \alpha +\ln(x)$ which is somewhat obvious but end up with a very similar result but with the an power not $\alpha$ but ($\alpha+1)$! This may be my algebra failing me ......but I keep arriving at this wrong result. Can someone please elaborate and guide me towards the correct answer?  
 A: Let $~u=\ln\left(x\right)+{\alpha}~$ , then $~ x~du= dx\implies dx=e^{u-\alpha}~du~$.
when $~x\to 1,~ u\to \alpha~$ and
when $~x\to e, ~u\to1+\alpha~$
Now
$$I=\int_{1}^{e}\frac{\ln(x)}{(\alpha +\ln(x))^{(\alpha+1)}}\,dx $$
$$=\class{steps-node}{\cssId{steps-node-1}{\mathrm{e}^{-{\alpha}}}}{\displaystyle\int_{\alpha}^{\alpha+1}}u^{-{\alpha}-1}\left(u-{\alpha}\right)\mathrm{e}^u\,\mathrm{d}u$$
$$=e^{-\alpha}~\int_{\alpha}^{\alpha+1}\left(\frac{e^u}{u^{\alpha}}-\alpha~e^u~u^{-{\alpha}-1}\right)du$$
$$=e^{-\alpha}~\int_{\alpha}^{\alpha+1}\frac{e^u}{u^{\alpha}}~du-\alpha ~e^{-\alpha}~\int_{\alpha}^{\alpha+1}~e^u~u^{-{\alpha}-1}~du$$
$$=e^{-\alpha}~\int_{\alpha}^{\alpha+1}\frac{e^u}{u^{\alpha}}~du-\alpha ~e^{-\alpha}~\left\{\left[-~\frac{1}{\alpha}~e^u~u^{-\alpha}\right]_{\alpha}^{\alpha+1}+~\frac{1}{\alpha}~\int_{\alpha}^{\alpha+1}~e^u~u^{-{\alpha}}~du\right\}$$
$$= ~e^{-\alpha}~\left[e^u~u^{-\alpha}\right]_{\alpha}^{\alpha+1}$$
$$= ~e^{-\alpha}~[e^{\alpha+1}~({\alpha+1})^{-\alpha}-e^\alpha~\alpha^{-\alpha}]$$
$$= \frac{e}{(\alpha+1)^\alpha} - \frac{1}{\alpha^\alpha}$$
A: Since $$\frac{d}{dx}\left[x(\alpha+\ln x)^{-\alpha}\right]=\frac{\ln x}{(\alpha+\ln x)^{\alpha+1}}$$(your substitution will get you there if you spot $\frac{d}{du}u^{-\alpha}e^u=(u^{-\alpha}-\alpha u^{-\alpha-1})e^u$, or you could use the required value of the integral to guess the antiderivative), your integral is$$\left[x(\alpha+\ln x)^{-\alpha}\right]_1^e=e(\alpha+1)^{-\alpha}-\alpha^{-\alpha},$$as required.
