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Integrating e^(square root (x))


Function

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Tried doing that, but expressing it with dt wasn't making it a bit simpler:

 

[math]t=\sqrt{x}[/math]

[math]\Leftrightarrow dt=\frac{dx}{2\sqrt{x}}[/math]

[math]\Leftrightarrow dx=2\sqrt{x}dt[/math]

 

[math]\int{\cdots}=\int{2e^{t}\sqrt{x}dt}=\int{2e^{t}tdt}[/math]

(stupid.. could've just replaced the x in dx by t²)

[math]\int{\cdots}=2\int{te^{t}dt}[/math]

Partial integration:

[math]\int{\cdots}=2\int{td\left(e^t\right)}[/math]

[math]\int{\cdots}=2\cdot \left(t\cdot e^t-\int{e^t dt}\right)[/math]

[math]\int{\cdots}=2\cdot\left(t\cdot e^t-e^t\right)+C[/math]

[math]\int{\cdots}=2te^{t}-2e^{t}=2e^{t}(t-1)+C[/math]

[math]\int{\cdots}=2e^{\sqrt{x}}\left(\sqrt{x}-1\right)+C[/math]

 

With [math]C[/math] any constant the result may differ from in comparison to other possible results.

Correct?

Edited by Function
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Glad to see it's correct. Can't say that I completely recognize the method the man uses in his video, but well.. Everyone has his own ways of solving problems, right..

Edited by Function
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  • 2 weeks later...

Let t = sqrt(x) So dx = 2.t.dt Integrate 2 . e^t . t . dt u = 2t dv = e^t dt -> du = 2dt v = e^t -> integral(2e^t t dt) = 2 t e^t - integral(2 e^t dt) = 2 t e^t - 2 e^t = e^t (t-1) . 2 , t = sqrt_x as above

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  • 2 weeks later...

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