# Help with solving indefinite integral

I am working on this problem, attempting to find the indefinite integral: $$\int9(\sqrt{2x})dx$$ 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!

• It's just some algebraic simplification. Your answer is the same as what you call "The Solution". – MathNewbie May 24 '15 at 6:54
• You had the $x$ outside the root, I fixed it for you. – Gregory Grant May 24 '15 at 6:55
• $9(2^{1\over 5})({5\over 6})=\frac{9\times 5}6 2^{1\over 5}=\frac{15}2 2^{1\over 5}=\frac{15}{2^{4\over 5}}$ – Claude Leibovici May 24 '15 at 7:00
• $2^{-1} 2^{1/5} = 2^{1/5-1} =2^{1/5-5/5}=2^{-4/5}$ – mvw May 24 '15 at 7:05

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.