Evaluate the integral. ∫x5x−1dx

The key in this situation is to adjust the variables. In particular, if you encounter a square root, you should either set it to something squared under the square root or convert it to something else $u=$ whichever is beneath it. We'll choose the latter in this instance.
Let $u=5x-1$.Then, because we know ${\displaystyle du=\frac{du}{ dx } dx }$, we see that $du=5dx$ which is equivalent to ${\displaystyle k dx =\frac{1}{d}u}$. We obtain by substituting these into the integral.
${\displaystyle \int x\sqrt{u}\cdot \frac{1}{5}du}$
This is a portion of what we want, but there is still a xx in there, and we like to completely switch to uu's. I mean, keep in mind that we set $u=5x-1$. Let's solve that for x in terms of u:
${\displaystyle 5x=u+1\Rightarrow x=\frac{u+1}{5}}$
By replacing this, we obtain
${\displaystyle \int \frac{u+1}{5}\sqrt{u}\cdot \frac{1}{5}du=\frac{1}{25}\int (u+1)\sqrt{u}du=\frac{1}{25}\int (u^{\frac{3}{2}}+u^{\frac{1}{2}})du}$