Integrate : 1/x+2 root of x

Let I = \int \frac{d x}{x + 2 \sqrt{x}} = \int \frac{d x}{\sqrt{x} (\sqrt{x} + 2)} put \sqrt{x} + 2 = t \Rightarrow \frac{1}{2 \sqrt{x}} d x = d t \Rightarrow \frac{d x}{\sqrt{x}} = 2 d t Now, I = 2 \int \frac{d t}{t (t - 2)} = 2 \int \frac{d t}{t^{2} - 2 t} = 2 \int \frac{d t}{t^{2} - 2 t + 1 - 1} = 2 \int \frac{d t}{{(t - 1)}^{2} - {(1)}^{2}} = 2 \times \frac{1}{2} \log |\frac{t - 1 - 1}{t - 1 + 1}| + C = \log |\frac{\sqrt{x} + 2 - 2}{\sqrt{x} + 2}| + C = \log |\frac{\sqrt{x}}{\sqrt{x} + 2}| + C

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