\documentclass[a4paper,12pt]{article}
\setlength\oddsidemargin{0pt} \setlength\evensidemargin{0pt}
\setlength\topmargin{0pt}
\begin{document}
\parindent=0pt
{\bf Exam Question

Topic: Reduction Formula}

Let $\displaystyle I_n(x)=\int_1^x t(\ln t)^n\, dt,$ where $x>0$
and $n$ is a non-negative integer.

(a) Show that $\displaystyle I_n(x)=\frac{x^2}{2}(\ln
x)^n-\frac{n}{2}I_{n-1}(x),$ when $n\ge1.$

(b) Find $I_0(x),\ I_1(x)$ and $I_2(x).$

(c) Find $\displaystyle \int_1^{\sqrt{\mathrm{e}}} t(\ln t)^4\,
dt.$

\vspace{0.5in}

{\bf Solution}

\begin{eqnarray*}
\mathrm{(a)}\ \ I_n(x)=\int_1^x t(\ln t)^n\, dt&=&
\left[\frac{t^2}{2}(\ln t)^n\right]_1^x - \int_1^x\frac{t}{2}n(\ln
t)^{n-1}\, dt\\ &=&\frac{x^2}{2}(\ln x)^n-\frac{n}{2}I_{n-1}(x)
\end{eqnarray*}

\begin{eqnarray*}
\mathrm{(b)}\ \  I_0(x)&=&\frac{x^2}{2}-\frac{1}{2}\\ I_1(x)&=&
\frac{x^2}{2}(\ln
x)+\left(\frac{-1}{2}\right)\left(\frac{x^2}{2}\right)
+\left(\frac{-1}{2}\right)\left(\frac{-1}{2}\right)=
\frac{x^2}{2}(\ln x)-\frac{x^2}{4}+\frac{1}{4}\\ I_2(x)&=&
\frac{x^2}{2}(\ln x)^2-\frac{2}{2}I_0(x)= \frac{x^2}{2}(\ln x)^2
-\frac{x^2}{2}(\ln x)+\frac{x^2}{4}-\frac{1}{4}
\end{eqnarray*}

\begin{eqnarray*}
\mathrm{(c)}\ \
I_n(\sqrt{\mathrm{e}})&=&\frac{\mathrm{e}}{2^n}-\frac{n}{2}I_{n-1}(\sqrt{\mathrm{e}})\
\ so\\
I_4(\sqrt{\mathrm{e}})&=&\frac{\mathrm{e}}{32}-2I_3(\sqrt{\mathrm{e}})
=
\frac{\mathrm{e}}{32}-2\left(\frac{\mathrm{e}}{16}-\frac{3}{2}I_2(\mathrm{e})\right)
= \frac{\mathrm{e}}{32}-\frac{\mathrm{e}}{8}+3I_2(\mathrm{e})\\
&=&\frac{\mathrm{e}}{32}-\frac{\mathrm{e}}{8}+3\left(\frac{\mathrm{e}}{8}
-\frac{\mathrm{e}}{4}+\frac{\mathrm{e}}{4}-\frac{1}{4}\right)=\frac{9\mathrm{e}}{32}-\frac{3}{4}.
\end{eqnarray*}


\end{document}
