Indefinite Integration

Evaluate: $\;$ $\displaystyle \int \sin^{-1} \left(\dfrac{2x}{1 + x^2}\right) \; dx$

Let $\;$ $I = \displaystyle \int \sin^{-1} \left(\dfrac{2x}{1 + x^2}\right) \; dx$

Let $\;$ $x = \tan \theta$

Then, $\;$ $dx = \sec^2 \theta \; d\theta$

$\begin{aligned} \therefore \; I & = \int \sin^{-1} \left(\dfrac{2 \tan \theta}{1 + \tan^2 \theta}\right) \sec^2 \theta \; d\theta \\\\ & = \int \sin^{-1} \left[\sin \left(2 \theta\right)\right] \sec^2 \theta \; d\theta \;\;\; \left[\text{Note: } \sin \left(2A\right) = \dfrac{2 \tan A}{1 + \tan^2 A}\right] \\\\ & = \int 2 \theta \; \sec^2 \theta \; d\theta \;\;\; \left[\text{Note: } \sin^{-1} \left[\sin \left(A\right)\right] = A\right] \\\\ & = 2 \left\{\theta \int \sec^2 \theta \; d\theta - \int \left[\dfrac{d}{d \theta} \left(\theta\right) \times \int \sec^2 d \theta\right] d \theta \right\} \\\\ & \left[\text{Note: Integration by parts: } \int u \cdot v \; dx = u \int v \; dx - \int \left\{\dfrac{du}{dx} \times \int v \; dx \right\} dx\right] \\\\ & = 2 \left\{\theta \; \tan \theta - \int \tan \theta \; d\theta \right\} + c \;\;\; \left[\text{Note: } \int \sec^2 x \; dx = \tan x + c\right] \\\\ & = 2 \; \theta \; \tan \theta - 2\log \left|\sec \theta\right| + c \;\;\; \left[\text{Note: } \int \tan x \; dx = \log \left|\sec x\right| + c\right] \\\\ & = 2 \; \theta \; \tan \theta - \log \left|\sec \theta\right|^2 + c \\\\ & = 2 \; \theta \; \tan \theta - \log \left|1 + \tan^2 \theta\right| + c \;\;\; \left[\text{Note: } 1 + \tan^2 x = \sec^2 x\right] \end{aligned}$

$\because \;$ $x = \tan \theta$ $\implies$ $\theta = \tan^{-1}x$

$\therefore \;$ $I = 2 \; x \; \tan^{-1}x - \log \left|1 + x^2\right| + c$