$\int \frac{d x}{1+\sqrt{x}}=$

Question

Aicharya · Accepted Answer

Let $\quad \mathrm{I}=\int \frac{\mathrm{dx}}{1+\sqrt{\mathrm{x}}}$ Put $\quad \sqrt{\mathrm{x}}=\mathrm{t} \Rightarrow \frac{1}{2 \sqrt{\mathrm{x}}} \mathrm{dx}=\mathrm{dt} \Rightarrow \mathrm{dx}=2 \mathrm{t} \mathrm{dt}$ $	herefore \mathrm{I}=\int \frac{2 \mathrm{t}}{1+\mathrm{t}} \mathrm{dt}$ $\quad=2 \int \frac{(\mathrm{t}+1)-1}{1+\mathrm{t}} \mathrm{dt}=2 \int \mathrm{dt}-2 \int \frac{\mathrm{dt}}{1+\mathrm{t}}$ $\quad=2 \mathrm{t}-2 \log |1+\mathrm{t}|=2 \sqrt{\mathrm{x}}-2 \log |1+\sqrt{\mathrm{x}}|+\mathrm{c}$

$\int \frac{d x}{1+\sqrt{x}}=$

Solution