The given function is \( \displaystyle (x+2) \sqrt{x-3} \) and we need to find the integration of this function with respect to \( x \).
Solution:
Let \( u = x \, -3 \). Then: \(x = u + 3 \quad \text{and} \quad dx = du \)
Substitute into the integral:
\( \begin{aligned} &\int (x + 2) \sqrt{x – 3} \, dx \\ &= \int (u + 3 + 2) \sqrt{u} \, du \\ &= \int (u + 5) u^{1/2} \, du \\ &= \int \left( u^{3/2} + 5u^{1/2} \right) du \\ &= \int u^{3/2} \, du + 5 \int u^{1/2} \, du \\ &= \frac{u^{5/2}}{5/2} + 5 \cdot \frac{u^{3/2}}{3/2} + C, \quad \text{C is a constant}\\ &= \frac{2}{5}u^{5/2} + \frac{10}{3}u^{3/2} + C \\ &= \frac{2}{15}u^{3/2} \left(3u + 25 \right) + C \\ &= \frac{2}{15}(x-3)^{3/2} \left(3(x-3) + 25 \right) + C \quad \left( u= x-3 \right) \\ &= \frac{2}{15}(x-3)^{3/2} \left(3x + 16 \right) + C \\ &= \frac{2}{15}\left(3x + 16 \right) (x-3)^{3/2} + C \\ \end{aligned} \)Thus, the integration of the given function is
\( \boxed{ \int (x + 2) \sqrt{x – 3} \, dx = \frac{2}{15}\left(3x + 16 \right) (x-3)^{3/2} + C } \)Please let me know in the comments if you find any errors in this solution.