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Question Number 151753 by Integrals last updated on 22/Aug/21

Commented by puissant last updated on 22/Aug/21
$K=∫(√x)e^(√x) dx u=(√x)→ u^2 =x → dx=2udu K=2∫u^2 e^u du { ((i=u^2 )),((j′=e^u )) :} ⇒ { ((i′=2u)),((j=e^u )) :} K=2[u^2 e^u ]−4∫ue^u du { ((i=u)),((j′=e^u )) :} ⇒ { ((i′=1)),((j=e^u )) :} ⇒ K=2u^2 e^u −4ue^u +4e^u +C K=2xe^(√x) −4(√x)e^(√x) +4e^(√x) +C..$
$K = \int \sqrt{x} e^{\sqrt{x}} d x$ $u = \sqrt{x} \to u^{2} = x \to d x = 2 u d u$ $K = 2 \int u^{2} e^{u} d u$ ${\begin{cases} i = u^{2} \\ j^{'} = e^{u} \end{cases} \Rightarrow {\begin{cases} i^{'} = 2 u \\ j = e^{u} \end{cases}$ $K = 2 [u^{2} e^{u}] - 4 \int {u e}^{u} d u$ ${\begin{cases} i = u \\ j^{'} = e^{u} \end{cases} \Rightarrow {\begin{cases} i^{'} = 1 \\ j = e^{u} \end{cases}$ $\Rightarrow K = 2 u^{2} e^{u} - 4 {u e}^{u} + 4 e^{u} + C$ $K = 2 {x e}^{\sqrt{x}} - 4 \sqrt{x} e^{\sqrt{x}} + 4 e^{\sqrt{x}} + C . .$
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