Evaluate the integral: I = integral

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

Step 1 · Substitute

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

$$ I = \int\frac{u+1}{u-1}\cdot2u\,du = 2\int\frac{u^2+u}{u-1}\,du. $$

Step 2 · Divide and integrate

$$ \frac{u^2+u}{u-1} = u+2+\frac2{u-1} \quad\Rightarrow\quad 2\int\left(u+2+\frac2{u-1}\right)du = u^2+4u+4\ln|u-1|+C. $$

Step 3 · Back-substitute

Since $u^2=x+1$, the constant $+1$ in $u^2+4u$ absorbs into $C$:

$$ I = x + 4\sqrt{x+1} + 4\ln\left|\sqrt{x+1}-1\right| + C. $$

Summary Table

Additional Notes

• The domain requires $x\geq-1,\ x\neq0$.
• The result was verified by differentiation: $\dfrac{d}{dx}\left[x+4\sqrt{x+1}+4\ln|\sqrt{x+1}-1|\right] = 1+\dfrac2{\sqrt{x+1}}+\dfrac2{\sqrt{x+1}(\sqrt{x+1}-1)} = 1+\dfrac2{\sqrt{x+1}-1} = \dfrac{\sqrt{x+1}+1}{\sqrt{x+1}-1}$, matching the integrand.