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This is the integral I've come across while solving the workbook: $$I=\int\frac{x}{\sqrt[4]{x^3(a-x)}}dx, a>0$$ My solution isn't the same as one in the workbook, so, please, tell me where I made a mistake:$$I=\int\frac{x^\frac{4}{4}}{x^\frac{3}{4}\sqrt[4]{a-x}}dx=\int\frac{\sqrt[4]{x}}{\sqrt[4]{a-x}}dx$$ Substitution: $x=\frac{1}{t}, t=\frac{1}{x}, dx=-\frac{1}{t^2}dt$ $$I=-\int\frac{\frac{1}{t^2}\sqrt[4]{\frac{1}{t}}}{\sqrt[4]{a-\frac{1}{t}}}dt=-\int\frac{\sqrt[4]{\frac{1}{t}}}{t^2\sqrt[4]{a-\frac{1}{t}}}dt=-\int\frac{\sqrt[4]{\frac{1}{t}}}{\sqrt[4]{at^8-t^7}}dt=-\int\sqrt[4]{\frac{1}{at^9-t^8}}dt=$$$$=-\int\frac{1}{\sqrt[4]{at^9-t^8}}dt=-\int\frac{1}{\sqrt[4]{at^9-t^8}}dt=-\int\frac{1}{t^2\sqrt{at-1}}dt$$ Substitution: $p=at-1, t=\frac{p+1}{a}, dp=adt, dt=\frac{dp}{a}$ $$I=-\int\frac{dp}{a\frac{(p+1)^2}{a^2}\sqrt{p}}=-a\int\frac{dp}{\sqrt{p}(p+1)^2}$$ Substitution: $z=\sqrt{p}, p=z^2, dz=\frac{1}{2\sqrt{p}}dp, \frac{dp}{\sqrt{p}}=2dz$ $$I=-2a\int\frac{dz}{(z^2+1)^2}$$ Substitution: $z=\tan{q}, dz=\frac{1}{\cos^2{q}}dq$ $$I=-2a\int\frac{dq}{\cos^2{q}(\tan^2{q}+1)^2}=-2a\int\cos^2{q}dq=-a\int{(\cos{2q}+1)dq}=$$$$=-aq-a\int{(\cos{2q})dq}$$ Substitution: $2q=n, 2dq=dn$ $$I=-aq-\frac{a}{2}\int(\cos{n})dn=-aq-\frac{a}{2}\sin{(n)}=$$$$=-a\arctan{\sqrt{\frac{a}{x}-1}}-\frac{a}{2}\sin{(2\arctan{\sqrt{\frac{a}{x}-1}}}+C$$ So, yeah, it took like 5 substitutions, but, I still got a different solution than in my workbook, which is: $$I=\frac{a(\frac{x}{a-x})^\frac{1}{4}}{1+\frac{x}{a-x}}+\frac{a}{4\sqrt{2}}\ln{\frac{(\frac{x}{a-x})^\frac{1}{2}+\sqrt{2}(\frac{x}{a-x})^\frac{1}{4}+1}{(\frac{x}{a-x})^\frac{1}{2}-\sqrt{2}(\frac{x}{a-x})^\frac{1}{4}+1}}+\frac{a}{2\sqrt{2}}\arctan{\frac{(\frac{x}{a-x})^\frac{1}{4}-1}{\sqrt{2}(\frac{x}{a-x})^\frac{1}{4}}}+C$$ What could have I done wrong?

Edit: This is the wolfram alpha solution: enter image description here

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    $\begingroup$ I must say, this is some tenacious integration. Also, there is a nice identity for $\sin(\arctan(x))$, which I thought could possibly help you. Check it out: wolframalpha.com/input/?i=sin%28arctan%28x%29%29 $\endgroup$ – Mankind Mar 15 '15 at 21:52
  • $\begingroup$ divide x on both den and num. to get $\frac{1}{x^{-1/4}(a-x)^{1/4}} dx$ then substitute $t = x+a/2$ and you get $\frac{1}{(a/2+t)^{-1/4}(a/2-t)^{1/4}}$ $\endgroup$ – mathreadler Mar 15 '15 at 23:02
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    $\begingroup$ Just because two things look different doesn't necessarily mean that they are different. As far as Wolfram Alpha is concerned, you can always use the Simplify[...] or FullSimplify[...] command. $\endgroup$ – Lucian Mar 16 '15 at 3:08

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