Evaluate the integral: I = integral dx/((sqrt(x+2)+1)

Evaluate the integral: $$ I = \int \frac{dx}{\left(\sqrt{x+2}+1\right)\sqrt{\sqrt{x+2}-1}}. $$

Step 1 · First substitution

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

$$ I = \int\frac{2u\,du}{(u+1)\sqrt{u-1}}. $$

Step 2 · Second substitution

Let $v=\sqrt{u-1}$, so $u=v^2+1$, $du=2v\,dv$, $u+1=v^2+2$:

$$ I = \int\frac{2(v^2+1)\cdot2v\,dv}{(v^2+2)\cdot v} = 4\int\frac{v^2+1}{v^2+2}\,dv. $$

Step 3 · Simplify and integrate

$$ \frac{v^2+1}{v^2+2} = 1-\frac1{v^2+2} \quad\Rightarrow\quad 4\int\left(1-\frac1{v^2+2}\right)dv = 4v - 2\sqrt2\arctan\left(\frac v{\sqrt2}\right)+C. $$

Step 4 · Back-substitute twice

$$ I = 4\sqrt{u-1} - 2\sqrt2\arctan\left(\frac{\sqrt{u-1}}{\sqrt2}\right) + C = 4\sqrt{\sqrt{x+2}-1} - 2\sqrt2\arctan\left(\sqrt{\frac{\sqrt{x+2}-1}2}\right) + C. $$

Summary Table

Additional Notes

• The domain requires $x\geq-1$, so that $\sqrt{x+2}\geq1$ and the inner square root is defined.
• The result was verified numerically at $x=7$ (where $u=3,\ v=\sqrt2$): the chain-rule derivative of the boxed expression evaluates to $1/(4\sqrt2)$, matching the integrand $1/[(3+1)\sqrt2]=1/(4\sqrt2)$ there.