Question

Evaluate \[ \int \sqrt{x+\sqrt{x^2+2}}\,dx \]

• A. \(\frac{3}{2}\left(x+\sqrt{x^2+2}\right)^{\frac{3}{2}}-2\left(x+\sqrt{x^2+2}\right)^{\frac{1}{4}}+C\)
• B. \(\frac{1}{3}\left(x+\sqrt{x^2+2}\right)^{\frac{3}{2}}-2\left(x+\sqrt{x^2+2}\right)^{\frac{1}{4}}+C\)
• C. \(\left(x+\sqrt{x^2+2}\right)^{-\frac{3}{2}}-2\left(x+\sqrt{x^2+2}\right)^{-\frac{1}{2}}+C\)
• D. \(\frac{\left(x+\sqrt{x^2+2}\right)^2-6}{3\sqrt{x+\sqrt{x^2+2}}}+C\)

Hint:

For integrals involving \[ x+\sqrt{x^2+a^2}, \] use the substitution \[ u=x+\sqrt{x^2+a^2}. \] Then express \(x\) in terms of \(u\), which simplifies the radical expression.

Answer 1

The Correct Option is D

Step 1: Put a suitable substitution.
Let \[ u=x+\sqrt{x^2+2}. \] Then, \[ u-x=\sqrt{x^2+2}. \] Squaring both sides, \[ (u-x)^2=x^2+2. \] \[ u^2-2ux+x^2=x^2+2. \] So, \[ u^2-2ux=2. \] \[ 2ux=u^2-2. \] Hence, \[ x=\frac{u^2-2}{2u}. \] This can be written as \[ x=\frac{1}{2}\left(u-\frac{2}{u}\right). \]

Step 2: Differentiate with respect to \(u\).
Differentiating, \[ \frac{dx}{du} = \frac{1}{2}\left(1+\frac{2}{u^2}\right). \] Therefore, \[ dx=\frac{1}{2}\left(1+\frac{2}{u^2}\right)du. \]

Step 3: Transform the integral.
Since \[ u=x+\sqrt{x^2+2}, \] we have \[ \sqrt{x+\sqrt{x^2+2}}=\sqrt{u}=u^{\frac{1}{2}}. \] Thus, \[ \int \sqrt{x+\sqrt{x^2+2}}\,dx = \int u^{\frac{1}{2}}\cdot \frac{1}{2}\left(1+\frac{2}{u^2}\right)du. \] \[ = \int \left(\frac{1}{2}u^{\frac{1}{2}}+u^{-\frac{3}{2}}\right)du. \]

Step 4: Integrate term by term.
Now, \[ \int \frac{1}{2}u^{\frac{1}{2}}\,du = \frac{1}{2}\cdot \frac{u^{\frac{3}{2}}}{\frac{3}{2}} = \frac{1}{3}u^{\frac{3}{2}}. \] Also, \[ \int u^{-\frac{3}{2}}\,du = \frac{u^{-\frac{1}{2}}}{-\frac{1}{2}} = -2u^{-\frac{1}{2}}. \] Therefore, \[ \int \sqrt{x+\sqrt{x^2+2}}\,dx = \frac{1}{3}u^{\frac{3}{2}}-2u^{-\frac{1}{2}}+C. \]

Step 5: Convert into the option form.
Taking \(u^{-\frac{1}{2}}\) common, \[ \frac{1}{3}u^{\frac{3}{2}}-2u^{-\frac{1}{2}} = \frac{u^2-6}{3\sqrt{u}}. \] Since \[ u=x+\sqrt{x^2+2}, \] we get \[ \frac{u^2-6}{3\sqrt{u}} = \frac{\left(x+\sqrt{x^2+2}\right)^2-6} {3\sqrt{x+\sqrt{x^2+2}}}. \]

Step 6: Final conclusion.
Hence, \[ \boxed{ \int \sqrt{x+\sqrt{x^2+2}}\,dx = \frac{\left(x+\sqrt{x^2+2}\right)^2-6} {3\sqrt{x+\sqrt{x^2+2}}}+C } \]