sjeikdom0

2020-10-20

Evaluate the integral. $\int x\sqrt{5x-1}dx$

unett

Skilled2020-10-21Added 119 answers

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 $du=\frac{du}{dx}dx$, we see that $du=5dx$ which is equivalent to $kdx=\frac{1}{d}u$. We obtain by substituting these into the integral.

$\int x\sqrt{u}\cdot \frac{1}{5}du$

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:

$5x=u+1\Rightarrow x=\frac{u+1}{5}$

By replacing this, we obtain

$\int \frac{u+1}{5}\sqrt{u}\cdot \frac{1}{5}du=\frac{1}{25}\int (u+1)\sqrt{u}du=\frac{1}{25}\int ({u}^{\frac{3}{2}}+{u}^{\frac{1}{2}})du$