Question

Evaluate: $\int \sqrt{x+\sqrt{x^2+2}}dx$

• A. $(\frac{x+\sqrt{x^2+2}}{3}{)}^{5/2}-2(x+\sqrt{x^4+2}{)}^{-3/2}+c$
• B. $\frac{2}{7}(x+\sqrt{x^2+2}{)}^{7/2}-2(x+\sqrt{x^2+2})+c$
• C. $\frac{(x+\sqrt{x^2+2}{)}^{3/2}}{3}-2(x+\sqrt{x^2+2}{)}^{-1/2}+c$
• D. $\frac{(x+\sqrt{x^2+2}{)}^{9/2}}{3}+2(x+\sqrt{x^2+2}{)}^{-5/2}+c$

Answer 1

The correct option is C $\frac{(x+\sqrt{x^2+2}{)}^{3/2}}{3}-2(x+\sqrt{x^2+2}{)}^{-1/2}+c$

Let $\sqrt{x+\sqrt{x^2+2}}=t^2$

$\therefore t^4-2t^{2}x+x^2=x^2+2$

$\therefore x=\frac{t^2}{2}-\frac{1}{t^2}$

$dx=(t+\frac{2}{t^3})dt$

Substituting these values makes the question easy to solve.

We get $\int t^{2}dt+\int \frac{2}{t^2}dt=\frac{t^3}{3}-\frac{2}{t}+c$

Re-substitute the value of t, we get

$\frac{(x+\sqrt{x^2+2}{)}^{3/2}}{3}-2(x+\sqrt{x^2+2}{)}^{-1/2}+c$