Step 1: Look at the integral.
We must find $\int\dfrac{1+x^2}{\sqrt{1-x^2}}\,dx$. The $\sqrt{1-x^2}$ in the bottom is a strong hint to use a trigonometric substitution.
Step 2: Substitute.
Let $x=\sin\theta$, so $dx=\cos\theta\,d\theta$ and $\sqrt{1-x^2}=\cos\theta$. The integral becomes \[ \int\frac{1+\sin^2\theta}{\cos\theta}\cos\theta\,d\theta=\int(1+\sin^2\theta)\,d\theta. \]
Step 3: Rewrite $\sin^2\theta$.
Use $\sin^2\theta=\tfrac{1-\cos2\theta}{2}$ so the integrand becomes \[ 1+\frac{1-\cos2\theta}{2}=\frac{3}{2}-\frac{\cos2\theta}{2}. \]
Step 4: Integrate.
\[ \int\left(\frac{3}{2}-\frac{\cos2\theta}{2}\right)d\theta=\frac{3}{2}\theta-\frac{\sin2\theta}{4}+c. \]
Step 5: Break the $\sin2\theta$ term.
Since $\sin2\theta=2\sin\theta\cos\theta$, the second term is $\tfrac{1}{2}\sin\theta\cos\theta$. So the answer is \[ \frac{3}{2}\theta-\frac{1}{2}\sin\theta\cos\theta+c. \]
Step 6: Return to $x$.
Recall $\theta=\sin^{-1}x$, $\sin\theta=x$, and $\cos\theta=\sqrt{1-x^2}$. So \[ \frac{3}{2}\sin^{-1}x-\frac{x}{2}\sqrt{1-x^2}+c. \]
\[ \boxed{\dfrac{3}{2}\sin^{-1}x-\dfrac{x}{2}\sqrt{1-x^2}+c} \]