Integrate the expression $$\frac{x^3}{(x^2 + 1)^3}$$
Show that $$\frac{dv}{du} - \frac{nv}{u} = u^n \frac{d}{du}(vu^{-n}).$$ By considering \(x\) as a function of \(y\), or otherwise, find and sketch the solution of the differential equation $$\frac{dy}{dx} = \frac{y}{3x - 2y},$$ which passes through the point \(x = 0\), \(y = 1\).
The value of \(y\) is given by \(y = a + c \ln y\), where \(c\) is small. Show that \(y\) is given approximately by $$y = a + c \ln a + \frac{c^2}{a} \ln a$$ and find the term in \(c^3\).
If $$I_n = \int_0^{\pi/2} \cos^n \theta \, d\theta,$$ find a recurrence relation for \(I_n\) and deduce that $$I_n I_{n-1} = \frac{\pi}{2n}$$ for all integers \(n > 1\).
If \(y = e^{-x}\sin(x\sqrt{3})\), prove that \begin{align} \frac{d^n y}{dx^n} = (-2)^n e^{-x} \sin(x\sqrt{3} - \frac{1}{3}n\pi) \end{align} Hence, or otherwise, show that \begin{align} y = \frac{\sqrt{3}}{2}\sum_{n=0}^{\infty}\left(\frac{(2x)^{3n+1}}{(3n+1)!} - \frac{(2x)^{3n+2}}{(3n+2)!}\right) \end{align}
Using the fact that \begin{align} \lim_{n\to\infty}\left(\frac{b-a}{n}\sum_{m=1}^{n}f(a+m[b-a]/n)\right) = \int_{a}^{b}f(x)dx \end{align} show that \begin{align} \lim_{n\to\infty}\left(\frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{n+n}\right) = \ln 2 \end{align} Evaluate \begin{align} \lim_{n\to\infty}\left(\frac{n}{n^2+1}+\frac{n}{n^2+4}+...+\frac{n}{n^2+n^2}\right) \end{align}
Suppose that \(f(n)\) is a polynomial with rational coefficients of degree \(k > 0\) in \(n\) where \(n\) is an integer. Show that the function \begin{align} g(n) = f(n) - f(n-1) \end{align} is a polynomial of degree \(k-1\). Show also that if \(f(n)\) has the form \(g(n)\cdot g(n+1)\cdot g(n+2)\) (for some polynomial \(g(n)\)), then \(g(n)\) is divisible by \(g(n) \cdot g(n+1)\). Evaluate the sum \begin{align} f(n) = \sum_{r=1}^{n} \frac{1}{r^3(r+1)^3(r^2+r+1)} \end{align} and hence show that \(f(n)\) is a perfect cube for all values of \(n\).
Using the substitution \(x = e^t\) or otherwise solve \begin{align} x^2\frac{d^2y}{dx^2} - 4x\frac{dy}{dx} + 6y = 6\ln x - 5 \text{ for } x > 0 \end{align} given \(y(1) = 0\) and \(y(e) = 2\).