Find $\int \frac{1-3x}{1+9x^2} \, \text{d}x$. Use the substitution $x = 2\sin\theta$, where $\theta$ is acute, to find $\int \frac{(x-1)^2}{\sqrt{4-x^2}} \, \text{d}x$. Find $\frac{\text{d}}{\text{d}x}(\sin^{-1}(2x^2))$. Hence solve $\int (2x\sin^{-1}(2x^2)) \, \text{d}x$.

1. Split the integrand into two separate terms to apply standard integration formulae.

$\int \frac{1-3x}{1+9x^2} \, \text{d}x = \int \frac{1}{1+9x^2} \, \text{d}x - \int \frac{3x}{1+9x^2} \, \text{d}x$

2. Integrate the first term using the standard formula for $\frac{1}{a^2+x^2}$.

$\int \frac{1}{1+9x^2} \, \text{d}x = \int \frac{1}{1+(3x)^2} \, \text{d}x = \frac{1}{3}\tan^{-1}(3x)$

3. Integrate the second term by manipulating it into the form $\frac{\mathrm{f}'(x)}{\mathrm{f}(x)}$ to obtain a natural logarithm.

$\int \frac{3x}{1+9x^2} \, \text{d}x = \frac{1}{6} \int \frac{18x}{1+9x^2} \, \text{d}x = \frac{1}{6}\ln(1+9x^2)$

4. Combine the integrated terms and append the constant of integration.

$\frac{1}{3}\tan^{-1}(3x) - \frac{1}{6}\ln(1+9x^2) + C$
(Note: Absolute value brackets are not required for the natural logarithm as $1+9x^2 > 0$ for all real $x$.)

1. Find the differential $\text{d}x$ in terms of $\text{d}\theta$.

Given $x = 2\sin\theta$,
$\frac{\text{d}x}{\text{d}\theta} = 2\cos\theta \implies \text{d}x = 2\cos\theta \, \text{d}\theta$

2. Substitute $x$ and $\text{d}x$ into the integral and simplify the denominator.

$\int \frac{(2\sin\theta - 1)^2}{\sqrt{4 - 4\sin^2\theta}} (2\cos\theta) \, \text{d}\theta$
$= \int \frac{4\sin^2\theta - 4\sin\theta + 1}{\sqrt{4(1-\sin^2\theta)}} (2\cos\theta) \, \text{d}\theta$
Since $\theta$ is acute, $\cos\theta > 0$, so $\sqrt{4\cos^2\theta} = 2\cos\theta$:
$= \int \frac{4\sin^2\theta - 4\sin\theta + 1}{2\cos\theta} (2\cos\theta) \, \text{d}\theta$
$= \int (4\sin^2\theta - 4\sin\theta + 1) \, \text{d}\theta$

3. Use the double angle identity to prepare the integrand for integration.

Using the identity $\cos 2\theta = 1 - 2\sin^2\theta$, we get $4\sin^2\theta = 2(1 - \cos 2\theta)$:
$\int (2 - 2\cos 2\theta - 4\sin\theta + 1) \, \text{d}\theta$
$= \int (3 - 2\cos 2\theta - 4\sin\theta) \, \text{d}\theta$

4. Integrate with respect to $\theta$.

$= 3\theta - \sin 2\theta + 4\cos\theta + C$

5. Convert the expression back to be in terms of $x$.

From $x = 2\sin\theta$, we have $\sin\theta = \frac{x}{2}$ and $\theta = \sin^{-1}\left(\frac{x}{2}\right)$.
Using a right-angled triangle, the adjacent side is $\sqrt{2^2 - x^2} = \sqrt{4-x^2}$, giving $\cos\theta = \frac{\sqrt{4-x^2}}{2}$.
We also need $\sin 2\theta = 2\sin\theta\cos\theta = 2\left(\frac{x}{2}\right)\left(\frac{\sqrt{4-x^2}}{2}\right) = \frac{x\sqrt{4-x^2}}{2}$.
Substituting these back into the integrated expression:
$3\sin^{-1}\left(\frac{x}{2}\right) - \frac{x\sqrt{4-x^2}}{2} + 4\left(\frac{\sqrt{4-x^2}}{2}\right) + C$
$= 3\sin^{-1}\left(\frac{x}{2}\right) + \left(2 - \frac{x}{2}\right)\sqrt{4-x^2} + C$
$= 3\sin^{-1}\left(\frac{x}{2}\right) + \frac{4-x}{2}\sqrt{4-x^2} + C$

1. Differentiate $\sin^{-1}(2x^2)$ with respect to $x$ using the chain rule.

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

2. Set up integration by parts for the given integral, utilising the derivative found in the first part.

Let $u = \sin^{-1}(2x^2)$ and $\frac{\text{d}v}{\text{d}x} = 2x$.
Then $\frac{\text{d}u}{\text{d}x} = \frac{4x}{\sqrt{1-4x^4}}$ and $v = x^2$.
Using the integration by parts formula $\int u \frac{\text{d}v}{\text{d}x} \, \text{d}x = uv - \int v \frac{\text{d}u}{\text{d}x} \, \text{d}x$:
$\int 2x\sin^{-1}(2x^2) \, \text{d}x = x^2 \sin^{-1}(2x^2) - \int x^2 \left( \frac{4x}{\sqrt{1-4x^4}} \right) \, \text{d}x$
$= x^2 \sin^{-1}(2x^2) - \int \frac{4x^3}{\sqrt{1-4x^4}} \, \text{d}x$

3. Evaluate the remaining integral using substitution to obtain the final answer.

Let $w = 1-4x^4$, then $\text{d}w = -16x^3 \, \text{d}x$.
The remaining integral becomes:
$\int \frac{4x^3}{\sqrt{1-4x^4}} \, \text{d}x = \int -\frac{1}{4} w^{-\frac{1}{2}} \, \text{d}w$
$= -\frac{1}{4} \left( \frac{w^{\frac{1}{2}}}{\frac{1}{2}} \right) = -\frac{1}{2}\sqrt{1-4x^4}$
Substituting this back into the integration by parts result:
$\int 2x\sin^{-1}(2x^2) \, \text{d}x = x^2 \sin^{-1}(2x^2) - \left( -\frac{1}{2}\sqrt{1-4x^4} \right) + C$
$= x^2 \sin^{-1}(2x^2) + \frac{1}{2}\sqrt{1-4x^4} + C$