$$ int _ { 0 } ^ { 2 } \frac { x + 2 } { sqrt { x ^ { 2 } + 4 x } } d x $$
$$ int _ { 0 } ^ { 2 } \frac { x + 2 } { sqrt { x ^ { 2 } + 4 x } } d x $$
Answer
Explanation:
Step1: Simplify the integrand
Let (u = x^{2}+4x), then (du=(2x + 4)dx=2(x + 2)dx), and ((x + 2)dx=\frac{1}{2}du). When (x = 0), (u=0); when (x = 2), (u=4 + 8=12). The integral (\int_{0}^{2}\frac{x + 2}{\sqrt{x^{2}+4x}}dx=\frac{1}{2}\int_{0}^{12}u^{-\frac{1}{2}}du).
Step2: Integrate
Using the power - rule for integration (\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C(n\neq - 1)), for (n=-\frac{1}{2}), we have (\frac{1}{2}\int_{0}^{12}u^{-\frac{1}{2}}du=\frac{1}{2}\left[2u^{\frac{1}{2}}\right]_{0}^{12}).
Step3: Evaluate the definite integral
(\frac{1}{2}\times2\left[u^{\frac{1}{2}}\right]{0}^{12}=\left[\sqrt{u}\right]{0}^{12}=\sqrt{12}-\sqrt{0}=2\sqrt{3}).
Answer:
(2\sqrt{3})