Challenging +1.8 Part (a) is a standard substitution with trigonometric identities leading to a known integral form. Part (b) is significantly more challenging, requiring a non-obvious rational substitution that transforms the logarithmic integrand in a clever way, combined with careful handling of limits and likely exploiting symmetry or special properties. The combination of techniques and the insight needed for part (b) elevates this above typical A-level questions, though it remains within reach for strong students familiar with AEA-level problem-solving.
4.(a)Use the substitution \(x = \sqrt { 3 } \tan u\) to show that
$$\int \frac { 1 } { 3 + x ^ { 2 } } \mathrm {~d} x = p \arctan ( p x ) + c$$
where \(p\) is a real constant to be determined and \(c\) is an arbitrary constant.
(b)Use the substitution \(x = \frac { 3 u + 3 } { u - 3 }\) to determine the exact value of \(I\) where
$$I = \int _ { - 3 } ^ { 1 } \frac { \ln ( 3 - x ) } { 3 + x ^ { 2 } } \mathrm {~d} x$$
giving your answer in simplest form.
\includegraphics[max width=\textwidth, alt={}, center]{a8e9db6b-dfad-4278-82d8-a8fa5ba61008-10_2264_47_314_1984}
4.(a)Use the substitution $x = \sqrt { 3 } \tan u$ to show that
$$\int \frac { 1 } { 3 + x ^ { 2 } } \mathrm {~d} x = p \arctan ( p x ) + c$$
where $p$ is a real constant to be determined and $c$ is an arbitrary constant.\\
(b)Use the substitution $x = \frac { 3 u + 3 } { u - 3 }$ to determine the exact value of $I$ where
$$I = \int _ { - 3 } ^ { 1 } \frac { \ln ( 3 - x ) } { 3 + x ^ { 2 } } \mathrm {~d} x$$
giving your answer in simplest form.\\
\includegraphics[max width=\textwidth, alt={}, center]{a8e9db6b-dfad-4278-82d8-a8fa5ba61008-10_2264_47_314_1984}
\hfill \mbox{\textit{Edexcel AEA 2024 Q4 [16]}}