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