Question

$\int \frac{xdx}{\sqrt{1+x^2+\sqrt{{(1+x^2)}^3}}}$ is equal to :

• A. $\frac{1}{2}\ln (1+\sqrt{1+x^2})+c$
• B. $2\sqrt{1+\sqrt{(1+x^2)}}+c$
• C. $2(1+\sqrt{1+x^2})+c$
• D. none of these

Answer 1

Answer: (B)

$2\sqrt{1+\sqrt{(1+x^2)}}+c$

Let $I=\int \frac{xdx}{\sqrt{1+x^2+\sqrt{(1+x^2{)}^3}}}$

$I=\int \frac{x}{\sqrt{1+x^2}}\frac{1}{\sqrt{1+\sqrt{1+x^2}}}dx$

Substituting $1+\sqrt{1+x^2}=t$ $[\frac{dt}{dx}=\frac{x}{\sqrt{1+x^2}}]$

$I=\int \frac{1}{\sqrt{t}}dt$

$I=2\sqrt{t}+c$

$I=2\sqrt{1+\sqrt{1+x^2}}+c$