Reverse an arctangent chain and use a principal sine substitution

The inner derivative is $\dfrac d{dx}(\tan^{-1}2x)=2/(1+4x^2)$. Hence the integral is $\frac12 e^{\tan^{-1}2x}+C$.

Let $x=\sin\theta$ and $dx=\cos\theta\,d\theta$. The stated interval gives $\cos\theta>0$, so $\sqrt{1-\sin^2\theta}=|\cos\theta|=\cos\theta$. The transformed integral is $\int\sin^2\theta\,d\theta=\frac12\int(1-\cos2\theta)d\theta=\frac12[\theta-\frac12\sin2\theta]+C$. Using $\theta=\sin^{-1}x$ and $\cos\theta=\sqrt{1-x^2}$ gives $\frac12[\sin^{-1}x-x\sqrt{1-x^2}]+C$. The printed triangle illustrates these ratios for positive x; the signed sine identity also covers negative x in the stated interval.