<s>[INST] Find $\int \frac{x^2}{2x-1}\mathrm{d}x$ Use the substitution $2x-1$ to find $\int  \frac{x^2}{2x-1}\mathrm{d}x.$
I started out by using the substitution to find $\mathrm{d}x$ in terms of $\mathrm{d}u$.
$$
\frac {\mathrm{d}u}{\mathrm{d}x} = 2 
$$
$$
\mathrm{d}x = \frac {1}{2}\mathrm{d}u
$$
Now the question looks like this:
$$
\int \! \frac{x^2}{u} \, \cdot \frac {1}{2}\mathrm{d}u
$$
I then found $x^2$ in terms of $u$:
$$
2x-1 = u
$$
$$
x = \frac {u+1}{2}
$$
$$
x^2 = \left(\frac {u+1}{2}\right)^2
$$
I then substituted this back into the equation.
$$
\int \! \frac {(\frac {u+1}{2})^2}{u} \cdot \frac {1}{2}\, \mathrm{d}u
$$
$$
\int \! \frac {(\frac {u+1}{2})^2}{2u}\, \mathrm{d}u
$$
From here I have tried expanding out the brackets, and using a substitution to put $2u$ on the top of the fraction, but neither method got the correct answer, and I have spent a long time on the question. Am I on the right track here, or should I be approaching this any differently? Thanks
 [/INST] HINT
Here it is another way to solve it: consider the following algebraic manipulation
\begin{align*}
\frac{x^{2}}{2x-1} & = \frac{1}{4}\times\frac{4x^{2}}{2x-1} = \frac{1}{4}\times\frac{(4x^{2}-1) + 1}{2x-1} = \frac{1}{4}\times\left[\frac{4x^{2}-1}{2x-1} + \frac{1}{2x-1}\right]\\\\
& = \frac{1}{4}\times\left[2x+1 + \frac{1}{2x-1}\right]
\end{align*}
Can you take it from here?
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