Question

integral of (2x-3)/(sqrt(4x-(x^2)))

Answer 1

Complete the square:
\[\begin{align*}
4x-x^2&=-(x^2-4x)\\
&=-(x^2-4x+4-4)\\
&=-\left((x-2)^2-4\right)\\
&=4-(x-2)^2
\end{align*}\]
\[\int\frac{2x-3}{\sqrt{4x-x^2}}~dx=\int\frac{2x-3}{\sqrt{4-(x-2)^2}}~dx\]
Let \(x-2=2\sin u\), then \(dx=2\cos u ~du\). This also gives you \(2x-3=4\sin u+1\).
\[\int\frac{(4\sin u+1)(2\cos u)}{\sqrt{4-(2\sin u)^2}}~du\\
\int\frac{(4\sin u+1)\cos u}{\sqrt{1-\sin^2u}}~du\\
\int\frac{(4\sin u+1)\cos u}{\sqrt{\cos^2u}}~du\\
\int(4\sin u+1)~du\]