When solving integrals with mixed functions like exponentials and rational functions, it is helpful to split the integral into separate parts. Look for substitution opportunities (like \( u = \sqrt{x} \)) to simplify the integrand. Additionally, remember that the exponential function, \( e^x \), integrates easily, and such terms typically appear straightforwardly in the result. Always verify your substitution and simplify step-by-step.
The integral to be evaluated is:
\[ I = \int e^x \left( \frac{2x + 1}{2\sqrt{x}} \right) dx. \]
The integrand simplifies as follows:
\[ \frac{2x + 1}{2\sqrt{x}} = \frac{2x}{2\sqrt{x}} + \frac{1}{2\sqrt{x}} = \sqrt{x} + \frac{1}{2\sqrt{x}}. \]
Substituting this simplification into the integral yields:
\[ I = \int e^x \left( \sqrt{x} + \frac{1}{2\sqrt{x}} \right) dx. \]
The integral can be split into two parts:
\[ I = \int e^x \sqrt{x} \, dx + \frac{1}{2} \int e^x \frac{dx}{\sqrt{x}}. \]
Using the substitution \( u = \sqrt{x} \), which implies \( x = u^2 \) and \( dx = 2u \, du \), we can transform both terms.
For the first integral term:
\[ \int e^x \sqrt{x} \, dx = \int e^x u \cdot 2u \, du = \int e^x u^2 \, du. \]
Applying the substitution \( u = \sqrt{x} \), this term becomes \( e^x u^2 = e^x \sqrt{x}. \)
For the second integral term:
\[ \frac{1}{2} \int e^x \frac{dx}{\sqrt{x}} = \frac{1}{2} \int e^x u^{-1} \cdot 2u \, du = \int e^x du = e^x. \]
Combining the results of both terms:
\[ I = e^x \sqrt{x} + e^x + C. \]
The final simplified form of the integral is:
\[ I = e^x \sqrt{x} + C. \]
If \[ \int (\sin x)^{-\frac{11}{2}} (\cos x)^{-\frac{5}{2}} \, dx \] is equal to \[ -\frac{p_1}{q_1}(\cot x)^{\frac{9}{2}} -\frac{p_2}{q_2}(\cot x)^{\frac{5}{2}} -\frac{p_3}{q_3}(\cot x)^{\frac{1}{2}} +\frac{p_4}{q_4}(\cot x)^{-\frac{3}{2}} + C, \] where \( p_i, q_i \) are positive integers with \( \gcd(p_i,q_i)=1 \) for \( i=1,2,3,4 \), then the value of \[ \frac{15\,p_1 p_2 p_3 p_4}{q_1 q_2 q_3 q_4} \] is ___________.