<s>[INST] Solve for $\int \sqrt{x}(\sqrt{x}-2x)^2 dx $ $\int \sqrt{x}(\sqrt{x}-2x)^2 dx $ so I solved this using U-substitution where $u= \sqrt{x}$ so my $du2\sqrt{x}=dx$ then it will be $2 \int u^2(u-2u^2)^2$ and just expand then distribute the $u^2$ so i got $\frac{8u^7}{7}-\frac{4u^6}{3}+\frac{2u^5}{5}+c$ then just substitue $\sqrt{x}$ to all of the $u$ is that right? My friend was debating with me with the $du=dx$ thing he said that it's suppose to be equal. Like in this equation where $\int x^2\sqrt{x^3-5}$ the $du=dx$ here is $du = 3x^2dx$ so it will be $\frac{du}{3}=x^2dx$. He keeps on debating that U substitution SHOULD be equal like in my second example. Where he has the $x^2dx$ really in the equation. Is this true? can't I really just equal it to $dx$ like  $u= \sqrt{x}$ $du2\sqrt{x}=dx$
 [/INST] Not answering your question, but another way to solve is with partial integration, is to observe that $$\sqrt{x}(\sqrt{x}-2x)^2=\sqrt{x}\left(\sqrt{x}(1-2\sqrt{x})\right)^2=\sqrt{x}\cdot x\cdot(\sqrt{x}-2x)^2=x\dfrac{(1-2\sqrt{x})^2}{\sqrt{x}}$$
Now $$\left((1-2\sqrt{x})^3\right)'=3(1-2\sqrt{x})^2\dfrac{-2}{2\sqrt{x}}=-3\dfrac{(1-2\sqrt{x})^2}{\sqrt{x}}$$ Hence the given integral can be written as $$-\dfrac{1}{3}\int x \cdot\left((1-2\sqrt{x})^3\right)'dx$$ and the result follows with partial integration.

Concerning your question, no your friend is not correct. There is no rule stipulating the equality that he insists on.
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