Evaluate the integral: I = integral dx/(1+ sqrt x)

Evaluate the integral: $$ I = \int \frac{dx}{1+\sqrt x}. $$

Step 1 · Substitute

Let $u=\sqrt x$, so $x=u^2$ and $dx=2u\,du$:

$$ I = \int\frac{2u\,du}{1+u} = 2\int\frac u{1+u}\,du. $$

Step 2 · Simplify by division

$$ \frac u{1+u} = \frac{(u+1)-1}{1+u} = 1-\frac1{1+u}. $$

Step 3 · Integrate

$$ 2\int\left(1-\frac1{1+u}\right)du = 2u - 2\ln|1+u| + C. $$

Step 4 · Back-substitute

Since $u=\sqrt x\geq0$, $1+u>0$, so the absolute value can be dropped:

$$ I = 2\sqrt x - 2\ln\left(1+\sqrt x\right) + C. $$

Summary Table

Additional Notes

• The domain requires $x\geq0$ for the square root to be real.
• The result can be checked by differentiating: $\dfrac{d}{dx}\left[2\sqrt x-2\ln(1+\sqrt x)\right] = \dfrac1{\sqrt x}-\dfrac1{\sqrt x(1+\sqrt x)} = \dfrac1{\sqrt x}\cdot\dfrac{\sqrt x}{1+\sqrt x} = \dfrac1{1+\sqrt x}$, matching the integrand exactly.