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

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

Step 1 · Substitute

Let $u=\sqrt{x-1}$, so $x=u^2+1$ and $dx=2u\,du$:

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

Step 2 · Simplify and integrate

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

Step 3 · Back-substitute

Since $\sqrt{x-1}\geq0$, $1+u>0$:

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

Summary Table

Additional Notes

• The domain requires $x\geq1$.
• The result was verified by differentiation: $\dfrac{d}{dx}\left[2\sqrt{x-1}-2\ln(1+\sqrt{x-1})\right] = \dfrac1{\sqrt{x-1}}-\dfrac1{\sqrt{x-1}(1+\sqrt{x-1})} = \dfrac1{1+\sqrt{x-1}}$, matching the integrand.