Question

integrate the function x/√(x + 4).dx , x>0

Answer 1

$\bf{I=\int{\frac{x}{\sqrt{x+4}}}\,dx}$

(x + 4) = t =>x = t - 4 --------(1)

differentiate both sides,

dx = dt -------(2)

put equations (1) and (2) in I,

I = $\bf{\int{\frac{t-4}{\sqrt{t}}}\,dt}$

= $\bf{\int{\frac{t}{\sqrt{t}}}\,dt-4\int{\frac{1}{\sqrt{t}}}\,dt}$

=$\bf{\int{t^{1/2}}\,dt-4\int{t^{-1/2}}\,dt}$

=$\bf{\frac{t^{1/2+1}}{1/2+1}-4\frac{t^{-1/2+1}}{-1/2+1}+C}$

=$\bf{\frac{2}{3}t\sqrt{t}-8\sqrt{t}+C}$

put t = (x + 4)

therefore, I = $\bf{\frac{2}{3}(x+4)\sqrt{x+4}-8\sqrt{x+4}+C}$