Question

The value of ${\int }_1^2\frac{1}{\left(x+1\right)\left(\sqrt{(x^2-1}\right)}dx$ is

• A. $1$
• B. $\frac{1}{\sqrt{3}}$
• C. $2\sqrt{3}$
• D. $-2\sqrt{3}$

Answer 1

The correct option is B

$\frac{1}{\sqrt{3}}$

Explanation for the correct option:

Compute the required value:

Given: ${\int }_1^2\frac{1}{\left(x+1\right)\left(\sqrt{(x^2-1}\right)}dx$

Let $x=\sec \left(t\right)$ , $dx=\sec \left(t\right)·\tan \left(t\right)·dt$

when $$\begin{gathered} x=1,t={\sec }^{-1}\left(1\right)=0 \\ x=2,t={\sec }^{-1}\left(2\right)=\frac{\mathrm{\pi }}{3} \end{gathered}$$

$$\begin{gathered} ={\int }_0^{\frac{\mathrm{\pi }}{3}}\frac{\sec \left(t\right)·\tan \left(t\right)}{\left(\sec \left(t\right)+1\right)\left(\sqrt{({\sec }^2\left(t\right)-1}\right)}dt \\ ={\int }_0^{\frac{\mathrm{\pi }}{3}}\frac{\sec \left(t\right)·\tan \left(t\right)}{\left(\sec \left(t\right)+1\right)\left(\sqrt{({\sec }^2\left(t\right)-1}\right)}dt\mathbf{[}\mathbf{\because }\mathbf{1}\mathbf{+}{\mathbf{tan}}^{\mathbf{2}}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}{\mathbf{sec}}^{\mathbf{2}}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{]} \\ ={\int }_0^{\frac{\mathrm{\pi }}{3}}\frac{\sec \left(t\right)·\tan \left(t\right)}{\left(\sec \left(t\right)+1\right)\left(\sqrt{{\tan }^2\left(t\right)}\right)}dt \\ ={\int }_0^{\frac{\mathrm{\pi }}{3}}\frac{\sec \left(t\right)·\tan \left(t\right)}{\left(\sec \left(t\right)+1\right)\left(\tan \left(t\right)\right)}dt \\ ={\int }_0^{\frac{\mathrm{\pi }}{3}}\frac{\sec \left(t\right)}{\left(\sec \left(t\right)+1\right)}dt \\ ={\int }_0^{\frac{\mathrm{\pi }}{3}}\frac{{\displaystyle \frac{1}{\cos \left(t\right)}}}{\left({\displaystyle \frac{1}{\cos \left(t\right)}}+1\right)}dt\mathbf{[}\mathbf{\because }\mathbf{as}\mathbf{}\mathbf{sec}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}\frac{\mathbf{1}}{\mathbf{cos}\mathbf{\left(}\mathbf{t}\mathbf{\right)}}\mathbf{]} \\ ={\int }_0^{\frac{\mathrm{\pi }}{3}}\frac{1}{\left(\cos \left(t\right)+1\right)}dt\mathbf{}\mathbf{\left[}\mathbf{\because }\mathbf{cos}\mathbf{\left(}\mathbf{2}\mathbf{t}\mathbf{\right)}\mathbf{+}\mathbf{1}\mathbf{=}\mathbf{2}{\mathbf{cos}}^{\mathbf{2}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{\right]} \\ ={\int }_0^{\frac{\mathrm{\pi }}{3}}\frac{1}{2{\cos }^2\left({\displaystyle \frac{t}{2}}\right)}dt \\ =\frac{1}{2}{\int }_0^{\frac{\mathrm{\pi }}{3}}{\sec }^2\left(\frac{t}{2}\right)dt\mathbf{[}\mathbf{\because }\mathbf{as}\mathbf{}\mathbf{\int }{\mathbf{sec}}^{\mathbf{2}}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{d}\mathbf{x}\mathbf{=}\mathbf{tan}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{]} \\ =\frac{1}{2}\times 2\times {\left|\tan \left(\frac{t}{2}\right)\right|}_0^{\frac{\mathrm{\pi }}{3}} \\ =\tan \left(\frac{\mathrm{\pi }}{6}\right)-\tan \left(0\right) \\ =\frac{1}{\sqrt{3}}-0 \\ =\frac{1}{\sqrt{3}} \end{gathered}$$

Hence, option $B$ is the correct answer.