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What technique should i use to integrate $$\int(4-x)^{1/4}e^{\frac{1}{2}x}dx?$$

Ive tried to use algebraic manipulation and integration by parts but it just became complicated.

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There is the obvious simplification of $u = 4-x$, and then maybe $u = v^4$. But I suspect that your actual goal is to compute a definite integral (probably from $0$ to $4$), which opens up a wide variety of other possibilities.... – Hurkyl Oct 25 '12 at 8:11
have you considered change of variables? – user31280 Oct 25 '12 at 8:13
itnegration by parts – ILoveMath Oct 25 '12 at 8:18
@Hurkyl Yes i am actually working on a definite integral from 0 to 4. – Mynameis Tiara Oct 25 '12 at 8:33

The change of variable $x=4-2z$ yields $$\int_0^4(4-x)^{1/4}\mathrm e^{x/2}\mathrm dx=2^{1/4}\mathrm e^2\int_0^2z^{1/4}\mathrm e^{-z}\mathrm dz=2^{1/4}\mathrm e^2\cdot\gamma(5/4,2), $$ where $\gamma$ is the lower incomplete gamma function.

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Hi. That was really helpful. I didnt know the existence of the lower gamma function until i read your answer. If i may ask another (believe me, i im hoping this is my last question about gamma) question about the incomplete gamma function. Part of my solution using incomplete gamma function led to the ff. $$-2^{0.25}e^{x/2}\frac{d}{dx}\int_0^xu^{0.25-1}e^{-u}du=-2^{0.25}\frac{d}{dx}e^‌​{x/2}\gamma(0.25,x)$$ I got stuck not knowing what i replace to the gamma expression for me to differentiate it. thanks again. – Mynameis Tiara Oct 29 '12 at 4:23
Not sure what you are asking but $(d/dx)\gamma(a,x)=x^{a-1}e^{-x}$. – Did Oct 29 '12 at 6:07

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