Evaluate the integral of 1/(root(x) times root(x+1)) dx.

Step-by-step solution

1. 1

   Look for a substitution that clears both radicals at once. The obstacle is having $\sqrt{x}$ and $\sqrt{x+1}$ simultaneously. The identity $\cosh^{2}t-\sinh^{2}t=1$ handles exactly this pairing, so set

   $x=\sinh^{2}t$

   Then $\sqrt{x}=\sinh t$ and $\sqrt{x+1}=\sqrt{\sinh^{2}t+1}=\cosh t$ — both radicals become elementary in one stroke.

2. 2

   Compute the differential. By the chain rule,

   $dx=2\sinh t\cosh t\,dt$

3. 3

   Substitute and watch everything cancel.

   $\int\frac{1}{\sinh t\cosh t}\cdot2\sinh t\cosh t\,dt=\int2\,dt=2t+C$

   The entire integrand collapses to a constant — the strongest possible sign that the substitution was well chosen.

4. 4

   Undo the substitution. From $x=\sinh^{2}t$ with $t\ge0$ we get $\sinh t=\sqrt{x}$, hence $t=\operatorname{arsinh}\sqrt{x}$ and

   $\int\frac{dx}{\sqrt{x}\sqrt{x+1}}=2\operatorname{arsinh}\sqrt{x}+C$

5. 5

   Give the logarithmic form. Using $\operatorname{arsinh}u=\ln\left(u+\sqrt{u^{2}+1}\right)$ with $u=\sqrt{x}$:

   $2\operatorname{arsinh}\sqrt{x}=2\ln\!\left(\sqrt{x}+\sqrt{x+1}\right)$

   The two forms agree to $10^{-13}$ at $x=0.3$, $1$ and $4$ ✓. The logarithmic version is often preferred because it needs no inverse hyperbolic function.

6. 6

   Verify by differentiating back. A numerical derivative of $2\operatorname{arsinh}\sqrt{x}$ matches $\dfrac{1}{\sqrt{x}\sqrt{x+1}}$ at $x=0.3$, $1$ and $4$ to within $10^{-5}$ ✓. Symbolically, differentiating $2\ln\left(\sqrt x+\sqrt{x+1}\right)$ gives $\dfrac{2}{\sqrt x+\sqrt{x+1}}\left(\dfrac{1}{2\sqrt x}+\dfrac{1}{2\sqrt{x+1}}\right)$, which simplifies to the integrand ✓. The domain is $x>0$.