Finding value of $\displaystyle \int\sqrt{x^2+\sqrt{1+x^2}}~~dx$
Try: Let $\displaystyle I =\int\sqrt{x^2+\sqrt{x^2+1}}~~dx$
Put $x=\tan \theta$ and $dx=\sec^2\theta ~~d\theta$
So $\displaystyle I =\int \sqrt{\sec \theta+\tan^2\theta}\cdot \sec^2 \theta ~~d\theta$
Using By parts, We have
$\displaystyle I =\sqrt{\sec \theta+\tan^2\theta}\cdot \tan \theta-\frac{1}{2}\int\frac{\sec \theta \tan \theta+2\tan \theta \sec^2\theta}{\sqrt{\sec \theta +\tan^2 \theta}}\cdot \tan \theta ~~d\theta$
Could some help me to solve it , Thanks