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

1. Simplify the logarithmic expression using laws of logarithms

$\ln\left(\frac{3x}{x^2+1}\right) = \ln(3x) - \ln(x^2+1)$

2. Differentiate the simplified expression term by term

$\frac{d}{dx}[\ln(3x) - \ln(x^2+1)] = \frac{1}{3x} \cdot 3 - \frac{1}{x^2+1} \cdot 2x$

3. Simplify the derivative

$= \frac{1}{x} - \frac{2x}{x^2+1}$

1. State the derivative from part (a)

From part (a),
$\frac{d}{dx} \ln\left(\frac{3x}{x^2+1}\right) = \frac{d}{dx} (\ln 3x - \ln(x^2+1)) = \frac{1}{x} - \frac{2x}{x^2+1}$

2. Integrate both sides with respect to $x$

$\int \left( \frac{1}{x} - \frac{2x}{x^2+1} \right) dx = \ln\left(\frac{3x}{x^2+1}\right) + c_1$
$\int \frac{1}{x} dx - \int \frac{2x}{x^2+1} dx = \ln\left(\frac{3x}{x^2+1}\right) + c_1$

3. Rearrange to find the required integral and simplify

$\int \frac{2x}{x^2+1} dx = \ln x - \ln\left(\frac{3x}{x^2+1}\right) + c_2$
$= \ln x - (\ln 3 + \ln x - \ln(x^2+1)) + c_2$
$= \ln(x^2+1) - \ln 3 + c_2$
$= \ln(x^2+1) + C$