Evaluate the integral. ∫x5x−1dx

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unett · Accepted Answer

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 $u = 5 x - 1$ .Then, because we know $d u = \frac{d u}{d x} d x$ , we see that $d u = 5 d x$ which is equivalent to $k d x = \frac{1}{d} u$ . We obtain by substituting these into the integral.
$\int x \sqrt{u} \cdot \frac{1}{5} d u$
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 $u = 5 x - 1$ . Let's solve that for x in terms of u:
$5 x = u + 1 \Rightarrow x = \frac{u + 1}{5}$
By replacing this, we obtain
$\int \frac{u + 1}{5} \sqrt{u} \cdot \frac{1}{5} d u = \frac{1}{25} \int (u + 1) \sqrt{u} d u = \frac{1}{25} \int (u^{\frac{3}{2}} + u^{\frac{1}{2}}) d u$

Evaluate the integral. int x sqrt(5x-1) dx

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