Show that the curve \(y=x-\ln(x^2+2x)\) has exactly one stationary point.

Find the \(x\)-coordinate of this point.

(a) Differentiate

\(\ln(x^3+3x^2)\)

with respect to \(x\), simplifying your answer.

(b) Hence find

\(\int \frac{x+2}{x(x+3)}\,dx.\)

It is given that

\(y=\frac{\ln(3x^2-5)}{2x+1},\qquad 3x^2\gt 5.\)

(a) Find the equation of the normal to the curve at the point where \(x=\sqrt2\).

(b) Hence find the approximate change in \(y\) as \(x\) increases from \(\sqrt2\) to \(\sqrt2+h\), where \(h\) is small.

Given that

\(y=\ln(\sin x+3\cos x),\qquad 0\lt x\lt \frac{\pi}{2},\)

(a) find \(\dfrac{dy}{dx}\),

(b) solve the equation \(\dfrac{dy}{dx}=-\frac12\).

Given that

\(y=\ln(1+\sin x),\qquad 0\lt x\lt\pi,\)

(a) find \(\dfrac{dy}{dx}\),

(b) find the exact value of \(\dfrac{dy}{dx}\) when \(x=\dfrac{\pi}{6}\), giving your answer in the form \(\dfrac1{\sqrt a}\), where \(a\) is an integer,

(c) solve the equation \(\dfrac{dy}{dx}=\tan x\).

(i) Differentiate

\((x^2+3)\ln(x^2+3)\)

with respect to \(x\).

(ii) Hence find

\(\int x\ln(x^2+3)\,dx.\)

The function \(y\) is defined by \(y=\dfrac{\ln(4x^2-1)}{x+2}\).

(i) State the values of \(x\) for which \(y\) is not defined.

(ii) Find \(\dfrac{dy}{dx}\).

(iii) Hence find the approximate increase in \(y\) as \(x\) increases from \(2\) to \(2+h\), where \(h\) is small.

(i) Write \(\ln\left(\dfrac{2x+1}{2x-1}\right)\) as the difference of two logarithms.

A curve has equation \(y=\ln\left(\dfrac{2x+1}{2x-1}\right)+4x\), for \(x\gt \dfrac12\).

(ii) Using your answer to part (i), show that \(\dfrac{dy}{dx}=\dfrac{ax^2+b}{4x^2-1}\), where \(a\) and \(b\) are integers.

(iii) Hence find the \(x\)-coordinate of the stationary point on the curve.

(iv) Determine the nature of this stationary point.

Given that \(y=\dfrac{\ln(3x^2+2)}{x^2+1}\), find the value of \(\dfrac{dy}{dx}\) when \(x=2\), giving your answer as \(a+b\ln14\), where \(a\) and \(b\) are fractions in their simplest form.