To evaluate the integral \( \int x \sqrt{5x - 3} \, dx \), we can use u-substitution.
Let \( u = 5x - 3 \), then \( du/dx = 5 \) or \( dx = du/5 \).
Now, substitute \( u \) and \( dx \) into the integral:
\[ \int x \sqrt{5x - 3} \, dx = \int x \sqrt{u} \, \frac{du}{5} \]
Now, we can pull the constant (1/5) out of the integral:
\[ \frac{1}{5} \int x \sqrt{u} \, du \]
Next, we can further simplify by expressing \( x \) in terms of \( u \) using \( u = 5x - 3 \), which gives \( x = \frac{u + 3}{5} \).
Substitute this into the integral:
\[ \frac{1}{5} \int \frac{u + 3}{5} \sqrt{u} \, du \]
Now, distribute the \( \frac{1}{5} \) into the expression:
\[ \frac{1}{25} \int (u + 3) \sqrt{u} \, du \]
Expand the expression:
\[ \frac{1}{25} \int (u^{3/2} + 3u^{1/2}) \, du \]
Now, integrate each term separately:
\[ \frac{1}{25} \left( \frac{2}{5}u^{5/2} + 2u^{3/2} \right) + C \]
Finally, substitute back \( u = 5x - 3 \) to get the final result:
\[ \frac{1}{25} \left( \frac{2}{5}(5x - 3)^{5/2} + 2(5x - 3)^{3/2} \right) + C \]
where \( C \) is the constant of integration.
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