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  $u= \sqrt{x}$ $du2\sqrt{x}=dx$
 A: expanding the integrand we obtain $\sqrt{x}(x-4x\sqrt{x}+4x^2)=x\sqrt{x}+4x^2\sqrt{x}-4x^2)$
the result in a few minutes
the antiderivative is given by $\frac{2}{105} x^{5/2} \left(60 x-70 \sqrt{x}+21\right)+C$
possible is also $u=\sqrt{x}$ thus we get $u^2=x$ and $dx=2udu$
A: 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.
A: Using your substitutiuon, $u = \sqrt x \implies du = \dfrac{dx}{2\sqrt x} \implies dx = 2\sqrt x \,du = 2u\,du$.
There's nothing wrong with this approach! 
But since you expanded anyway, after u-substitution, you may have well done that from the beginning, in which case you wouldn't have had to substitute or express $dx$ in terms of $du$.
And for the record, your solution is correct, provided you back-substitute $u = \sqrt x$ into your answer expressed as a polynomial in $u$. 
Note: What is nice about your method is that in dealing strictly with the integration of a polynomial, one might be less prone to silly arithmetic errors that can sometimes  result when integrating expressions with fractional exponents.
