Find

1. Rewrite the integrand using the trigonometric identity $\tan^2 \theta = \sec^2 \theta - 1$.

$\int \tan^2(x - 1) \text{ d}x = \int (\sec^2(x - 1) - 1) \text{ d}x$

2. Integrate term-by-term and add the constant of integration.

$= \tan(x - 1) - x + C$

1. Set up integration by parts by identifying $u$ and $\frac{dv}{dx}$.

$\int \sin^{-1} 2x \text{ d}x = \int 1 \cdot \sin^{-1} 2x \text{ d}x$
Let $u = \sin^{-1} 2x \implies \frac{du}{dx} = \frac{2}{\sqrt{1 - (2x)^2}} = \frac{2}{\sqrt{1 - 4x^2}}$
Let $\frac{dv}{dx} = 1 \implies v = x$

2. Apply the integration by parts formula $\int u \text{ d}v = uv - \int v \text{ d}u$.

$\int \sin^{-1} 2x \text{ d}x = x \sin^{-1} 2x - \int x \left( \frac{2}{\sqrt{1 - 4x^2}} \right) \text{ d}x$
$= x \sin^{-1} 2x - \int \frac{2x}{\sqrt{1 - 4x^2}} \text{ d}x$

3. Evaluate the remaining integral using reverse chain rule (substitution) and add the constant of integration.

Consider $\int \frac{2x}{\sqrt{1 - 4x^2}} \text{ d}x$:
$= -\frac{1}{4} \int -8x (1 - 4x^2)^{-\frac{1}{2}} \text{ d}x$
$= -\frac{1}{4} \left( \frac{(1 - 4x^2)^{\frac{1}{2}}}{\frac{1}{2}} \right) = -\frac{1}{2}\sqrt{1 - 4x^2}$
Substituting this back:
$\int \sin^{-1} 2x \text{ d}x = x \sin^{-1} 2x - \left( -\frac{1}{2}\sqrt{1 - 4x^2} \right) + C$
$= x \sin^{-1} 2x + \frac{1}{2}\sqrt{1 - 4x^2} + C$