$\int \frac{x}{\sqrt{1+x^2+\sqrt{\left(1+x^2\right)^3}}} d x$ is equal to
Answer & explanation
Correct answer: option 4
Let $1+x^2=t^2$. Then, $x d x=t d t$
∴ $I=\int \frac{x}{\sqrt{1+x^2+\sqrt{\left(1+x^2\right)^3}}} d x$
$\Rightarrow I=\int \frac{t}{\sqrt{t^2+t^3}} d t=\int \frac{1}{\sqrt{1+t}} d t=2 \sqrt{1+t}+C$
$\Rightarrow I=2 \sqrt{1+\sqrt{1+x^2}}+C$