Question

Answer 1

$$\begin{gathered} \int \frac{dx}{1+\sqrt{x}} \\ =\int \frac{\sqrt{x}dx}{\sqrt{x}\left(1+\sqrt{x}\right)} \\ \mathrm{Let}1+\sqrt{x}=t \\ \Rightarrow \sqrt{x}=t-1 \\ \Rightarrow \frac{1}{2\sqrt{x}}=\frac{dt}{dx} \\ \Rightarrow \frac{dx}{\sqrt{x}}=2dt \end{gathered}$$

$$\begin{gathered} \mathrm{Now},\int \frac{\sqrt{x}}{\sqrt{x}\left(1+\sqrt{x}\right)}dx \\ =2\int \left(\frac{t-1}{t}\right)dt \\ =2\int \left(1-\frac{1}{t}\right)dt \\ =2\left(t-\log \left|t\right|\right)+\mathrm{C} \\ =2\left(1+\sqrt{x}\right)-2\log \left|1+\sqrt{x}\right|+\mathrm{C} \\ \mathrm{Let}\mathrm{C}+2=\mathrm{C}\text{'} \\ =2\sqrt{x}-2\log \left(1+\sqrt{x}\right)+\mathrm{C}\text{'} \end{gathered}$$