Given: \[ \int x e^{\sqrt{x}} \, dx \] Let's solve this step-by-step using **substitution** and **integration by parts**. --- ## **Step 1: Substitution** Let \( u = \sqrt{x} \implies x = u^2 \). Now, \( dx = 2u \, du \). --- ## **Step 2: Substitute into the Integral** \[ \int x e^{\sqrt{x}} \, dx = \int u^2 e^{u} \cdot 2u \, du = 2 \int u^3 e^{u} \, du \] --- ## **Step 3: Integration by Parts** Let’s use integration by parts for \( \int u^3 e^{u} \, du \): Let - \( a = u^3 \implies da = 3u^2 du \) - \( db = e^u du \implies b = e^u \) By parts: \[ \int a\, db = ab - \int b\, da \] \[ \int u^3 e^u du = u^3 e^u - \int e^u \cdot 3u^2 du \] Now, repeat integration by parts for \( \int u^2 e^u du \): \[ \int u^2 e^u du = u^2 e^u - \int e^u \cdot 2u du \] \[ = u^2 e^u - 2 \int u e^u du \] Now, integrate \( \int u e^u du \): \[ \int u e^u du = u e^u - \int e^u du = u e^u - e^u \] Plug back: \[ \int u^2 e^u du = u^2 e^u - 2 (u e^u - e^u) = u^2 e^u - 2u e^u + 2e^u \] Now, substitute back into the first by-parts: \[ \int u^3 e^u du = u^3 e^u - 3(u^2 e^u - 2u e^u + 2e^u) \] \[ = u^3 e^u - 3u^2 e^u + 6u e^u - 6 e^u \] \[ = (u^3 - 3u^2 + 6u - 6) e^u \] --- ## **Step 4: Substitute Back \(u = \sqrt{x}\) and Multiply by 2** Recall, the original integral is \( 2 \int u^3 e^u du \): So, \[ 2 (u^3 - 3u^2 + 6u - 6) e^u + C \] Substitute \( u = \sqrt{x} \): \[ 2\left((\sqrt{x})^3 - 3(\sqrt{x})^2 + 6(\sqrt{x}) - 6\right) e^{\sqrt{x}} + C \] \[ = 2\left(x^{3/2} - 3x + 6x^{1/2} - 6\right) e^{\sqrt{x}} + C \] --- ## **Final Answer** \[ \boxed{ \int x e^{\sqrt{x}} \, dx = 2\left( x^{3/2} - 3x + 6x^{1/2} - 6 \right) e^{\sqrt{x}} + C } \]