To solve the integral \(\int \sqrt{x} e^{\sqrt{x}} \, dx\), we can use the substitution method. Let \( u = \sqrt{x} \), then \( du = \frac{1}{2\sqrt{x}} dx \) or \( dx = 2u \, du \). Substituting these into the integral simplifies it to a form that is easier to integrate.
To solve the integral \(\int \sqrt{x} e^{\sqrt{x}} \, dx\), we start by using the substitution method. Let \( u = \sqrt{x} \), then \( du = \frac{1}{2\sqrt{x}} dx \) or \( dx = 2u \, du \).
Substituting \( u \) and \( dx \) into the integral, we get:
\[
\int \sqrt{x} e^{\sqrt{x}} \, dx = \int u e^u \cdot 2u \, du = 2 \int u^2 e^u \, du
\]
To integrate \( 2 \int u^2 e^u \, du \), we use integration by parts. Let \( v = u^2 \) and \( dw = e^u \, du \). Then \( dv = 2u \, du \) and \( w = e^u \).
Using the integration by parts formula \(\int v \, dw = vw - \int w \, dv\), we get:
\[
2 \int u^2 e^u \, du = 2 \left( u^2 e^u - \int 2u e^u \, du \right)
\]
Now, we need to integrate \( \int 2u e^u \, du \). Again, using integration by parts with \( v = 2u \) and \( dw = e^u \, du \), we get:
\[
\int 2u e^u \, du = 2u e^u - \int 2 e^u \, du = 2u e^u - 2 e^u
\]
Substituting back, we have:
\[
2 \left( u^2 e^u - \left( 2u e^u - 2 e^u \right) \right) = 2u^2 e^u - 4u e^u + 4 e^u
\]
Finally, substituting \( u = \sqrt{x} \) back into the expression, we get:
\[
2 (\sqrt{x})^2 e^{\sqrt{x}} - 4 \sqrt{x} e^{\sqrt{x}} + 4 e^{\sqrt{x}} = 2x e^{\sqrt{x}} - 4 \sqrt{x} e^{\sqrt{x}} + 4 e^{\sqrt{x}}
\]
\(\boxed{2x e^{\sqrt{x}} - 4 \sqrt{x} e^{\sqrt{x}} + 4 e^{\sqrt{x}} + C}\)
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