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$$ \int \frac{7x^5}{\sqrt{5-x^3}}\mathrm dx $$

Can someone teach me about the best solution to solve this integration? I know I can't use substitution here. Is there a formula for this kind of integration?

I'm definitely blank about this one. Thanks

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Why do you think you can't use substitution? – Daniel Fischer Sep 10 '13 at 11:08
well somehow when looking at the equation, it just didn't occur in my mind to use substitution. but yeah, I guess I was wrong. :\ – Karate_Dog Sep 10 '13 at 12:37
Ah, just a "didn't see how". That's normal. I thought it might be something deeper, a misconception that needed rectification, that's why I asked. – Daniel Fischer Sep 10 '13 at 12:44
up vote 6 down vote accepted

Substitute $u=x^3$:

$$7 \int dx \frac{x^5}{\sqrt{5-x^3}} = \frac{7}{3} \int d(x^3) \frac{x^3}{\sqrt{5-x^3}} = \frac{7}{3} \int du \, u (5-u)^{-1/2}$$

Now integrate by parts:

$$\frac{7}{3} \int du \, u (5-u)^{-1/2} = -\frac{14}{3} u (5-u)^{1/2} + \frac{14}{3} \int du \, (5-u)^{1/2} $$

which I assume you can take from here.

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You may consult the method Differential Binomial to use another substitution $$5-x^3=t^2$$

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Nice alternate hint! =1 – Amzoti Sep 10 '13 at 11:59
Short and sweet! – amWhy Sep 10 '13 at 14:03
@amWhy: Good night Amy. Have a colorful sleep. :-) – Babak S. Sep 11 '13 at 1:52

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