Integral of $f(x,\lambda)=\frac{x}{\sqrt{x-\lambda x^2}}$

Question

How can I compute the indefinite integral:

\begin{equation} I=\int \frac{x}{\sqrt{x-\lambda x^2}}dx \end{equation} with $\lambda$ a positive real parameter? Thank you!

Answer 1

By completing the square, $x-\lambda x^2=1/(4\lambda^2)-(x-\lambda/2)^2$, then use a trigonometric substitution.