Question

The value of the integral $\int_{0}^{\frac{1}{\sqrt{3}}} \sqrt{x+\sqrt{x^2+1}} dx$ is

• A. 1
• B. 2/3
• C. 2
• D. 3/2

Answer 1

Answer: (B)

2/3

Answer 2

To evaluate the integral $I = \int_{0}^{\frac{1}{\sqrt{3}}} \sqrt{x+\sqrt{x^2+1}} dx$, we use a hyperbolic substitution.

Let $x = \sinh t$.
Then $dx = \cosh t \, dt$.
Also, $\sqrt{x^2+1} = \sqrt{\sinh^2 t + 1} = \sqrt{\cosh^2 t} = \cosh t$ (since $\cosh t > 0$).

Substitute these into the integrand:
$\sqrt{x+\sqrt{x^2+1}} = \sqrt{\sinh t + \cosh t}$.
We know that $\sinh t + \cosh t = e^t$.
So, the integrand becomes $\sqrt{e^t} = e^{t/2}$.

Now, we need to change the limits of integration:

1. When $x=0$:
   $\sinh t = 0 \implies t=0$.
2. When $x=\frac{1}{\sqrt{3}}$:
   $\sinh t = \frac{1}{\sqrt{3}}$.
   Using the formula $\text{arsinh}(y) = \ln(y+\sqrt{y^2+1})$:
   $t = \ln\left(\frac{1}{\sqrt{3}} + \sqrt{\left(\frac{1}{\sqrt{3}}\right)^2+1}\right)$
   $t = \ln\left(\frac{1}{\sqrt{3}} + \sqrt{\frac{1}{3}+1}\right)$
   $t = \ln\left(\frac{1}{\sqrt{3}} + \sqrt{\frac{4}{3}}\right)$
   $t = \ln\left(\frac{1}{\sqrt{3}} + \frac{2}{\sqrt{3}}\right)$
   $t = \ln\left(\frac{3}{\sqrt{3}}\right) = \ln(\sqrt{3})$.

So the integral transforms to:
$I = \int_{0}^{\ln(\sqrt{3})} e^{t/2} \cosh t \, dt$.

Now, express $\cosh t$ in terms of exponentials: $\cosh t = \frac{e^t + e^{-t}}{2}$.
$I = \int_{0}^{\ln(\sqrt{3})} e^{t/2} \left(\frac{e^t + e^{-t}}{2}\right) dt$
$I = \frac{1}{2} \int_{0}^{\ln(\sqrt{3})} (e^{t/2} e^t + e^{t/2} e^{-t}) dt$
$I = \frac{1}{2} \int_{0}^{\ln(\sqrt{3})} (e^{3t/2} + e^{-t/2}) dt$.

Now, integrate term by term:
$\int e^{3t/2} dt = \frac{e^{3t/2}}{3/2} = \frac{2}{3} e^{3t/2}$.
$\int e^{-t/2} dt = \frac{e^{-t/2}}{-1/2} = -2 e^{-t/2}$.

So, the definite integral is:
$I = \frac{1}{2} \left[ \frac{2}{3} e^{3t/2} - 2 e^{-t/2} \right]_{0}^{\ln(\sqrt{3})}$
$I = \left[ \frac{1}{3} e^{3t/2} - e^{-t/2} \right]_{0}^{\ln(\sqrt{3})}$.

Now, evaluate at the limits:
At the upper limit $t=\ln(\sqrt{3})$:
$e^{3t/2} = e^{\frac{3}{2}\ln(\sqrt{3})} = e^{\ln((\sqrt{3})^{3/2})} = (\sqrt{3})^{3/2} = (3^{1/2})^{3/2} = 3^{3/4}$.
$e^{-t/2} = e^{-\frac{1}{2}\ln(\sqrt{3})} = e^{\ln((\sqrt{3})^{-1/2})} = (\sqrt{3})^{-1/2} = (3^{1/2})^{-1/2} = 3^{-1/4}$.

So, at $t=\ln(\sqrt{3})$:
$\frac{1}{3} (3^{3/4}) - 3^{-1/4} = 3^{-1} 3^{3/4} - 3^{-1/4} = 3^{-1+3/4} - 3^{-1/4} = 3^{-1/4} - 3^{-1/4} = 0$.

At the lower limit $t=0$:
$\frac{1}{3} e^{3(0)/2} - e^{-(0)/2} = \frac{1}{3} e^0 - e^0 = \frac{1}{3}(1) - 1 = \frac{1}{3} - 1 = -\frac{2}{3}$.

Finally, subtract the lower limit value from the upper limit value:
$I = 0 - \left(-\frac{2}{3}\right) = \frac{2}{3}$.