Let \(u =x^2 + 4\)
\(du = 2xdx\)
\(\frac 12 du = xdx\)
\(\Rightarrow \int \frac x{\sqrt {x^2 + 4}} dx = \frac12 \int \frac 1 {\sqrt u} du\)
\(= \frac 12 \int \frac1{u^{1/2}} du\)
\(= \frac 12 \int u^{-1/2} du\)
Use power rule \(\int x^n dx = \frac{x^{n+1}}{n + 1}\) where \(n \ne 1\)
\(= \frac12 \left(\frac{u^{-\frac 12 + 1}}{- \frac12 +1}\right)\)
\(= \frac 12 \left( \frac{u^{1/2}}{1/2}\right)\)
Substitute back \(u = x^2+4\), we get
\(= (x^2 + 4)^{1/2}\)
\(= \sqrt{x^2 + 4 }+ C\)
