I am working on this problem, attempting to find the indefinite integral: $$\int9(\sqrt[5]{2x})dx$$ I can manage to get up to here: $$=9(2^{1\over 5})({5\over 6}x^\frac{6}{5})+C$$ But I don't know how to get to here (The Solution): $$=\frac{15x^\frac{6}{5}}{2^{4\over 5}}+C$$ If anyone could provide an explanation as to how I get from where I am, to the solution, it would be greatly appreciated!
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6It's just some algebraic simplification. Your answer is the same as what you call "The Solution". – MathNewbie May 24 '15 at 06:54
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You had the $x$ outside the root, I fixed it for you. – Gregory Grant May 24 '15 at 06:55
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2$9(2^{1\over 5})({5\over 6})=\frac{9\times 5}6 2^{1\over 5}=\frac{15}2 2^{1\over 5}=\frac{15}{2^{4\over 5}}$ – Claude Leibovici May 24 '15 at 07:00
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$2^{-1} 2^{1/5} = 2^{1/5-1} =2^{1/5-5/5}=2^{-4/5}$ – mvw May 24 '15 at 07:05
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As a gentle reminder, please note that, by the first fundamental theorem of calculus, the indefinite integral of a function is a primitive of the function. So it is better to call $\int f$ a primitive of a function $f$.
Regarding the original question, we have $\int 9\cdot 2^{1/5}\cdot x^{1/5} dx = 15\cdot 2^{-4/5}\cdot x^{6/5} +$ const.
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write 6=2.3 then get cancel nominator 9 by denominator 3 and multiply it by present 5 in nominator you will get 15. And then handelout 2 by (2)^1/5 you will get (2)^4/5 in the denominator
