Differentiate $\ln\left(1+3x^2\right)^2$ with respect to $x$. Hence, find $\displaystyle \int \frac{6x}{1+3x^2}\, dx$.

1. Simplify the expression using the power law of logarithms

Let $y = \ln\left(1+3x^2\right)^2$.
Using the power law of logarithms, $y = 2\ln(1+3x^2)$.

2. Differentiate the simplified expression using the chain rule

$\frac{dy}{dx} = 2 \cdot \frac{1}{1+3x^2} \cdot \frac{d}{dx}(1+3x^2)$
$\frac{dy}{dx} = 2 \cdot \frac{1}{1+3x^2} \cdot (6x)$
$\frac{dy}{dx} = \frac{12x}{1+3x^2}$

1. Recall the derivative from part (i)

From (i), $\frac{d}{dx} \left[ \ln(1+3x^2)^2 \right] = \frac{12x}{1+3x^2}$.

2. Use integration as the reverse of differentiation

$\int \frac{12x}{1+3x^2}\, dx = \ln(1+3x^2)^2 + c_1$

$\int \frac{6x}{1+3x^2}\, dx = \frac{1}{2} \int \frac{12x}{1+3x^2}\, dx$

$= \frac{1}{2} \ln(1+3x^2)^2 + c$

$= \ln(1+3x^2) + c$