Year: 2025
Paper: 2
Question Number: 5
Course: LFM Pure
Section: Integration
As is commonly the case, the vast majority of candidates focused on the Pure questions in Section A of the paper, with a good number of attempts made on all of those questions. Candidates that attempted the Mechanics questions in Section B generally answered both questions. More candidates attempted Question 11 in Section C than either Mechanics question, but very few attempted Question 12 in that section. There were a large number of good responses seen for all the questions, but a significant number of responses lacked sufficient detail in the presentation, particularly when asked to prove a given result or provide an explanation. Candidates who did well on this paper generally: gave careful explanations of each step within their solutions; indicated all points of interest on graphs and other diagrams clearly; made clear comments about the approach that needed to be taken, particularly when having to explore a number of cases as part of the solution to a question; used mathematical terminology accurately within their solutions. Candidates who did less well on this paper generally: made errors with basic algebraic manipulation, such as incorrect processing of indices; produced sketches of graphs in which significant points were difficult to see clearly because of the chosen scale; skipped important lines within lengthy sections of algebraic reasoning.
Difficulty Rating: 1500.0
Difficulty Comparisons: 0
Banger Rating: 1500.0
Banger Comparisons: 0
You need not consider the convergence of the improper integrals in this question.
\begin{questionparts}
\item Use the substitution $x = u^{-1}$ to show that
\[\int_0^{\infty} \frac{\sqrt{x}-1}{\sqrt{x(x^3+1)}} \, dx = 0.\]
\item Use the substitution $x = u^{-2}$ to show that
\[\int_0^{\infty} \frac{1}{\sqrt{x^3+1}} \, dx = 2\int_0^{\infty} \frac{1}{\sqrt{x^6+1}} \, dx.\]
\item Find, in terms of $p$ and $s$, a value of $r$ for which
\[\int_0^{\infty} \frac{x^r - 1}{\sqrt{x^s(x^p+1)}} \, dx = 0,\]
given that $p$ and $s$ are fixed values for which the required integrals converge.
\item Show that, for any positive value of $k$, it is possible to find values of $p$ and $q$ for which
\[\int_0^{\infty} \frac{1}{\sqrt{x^p+1}} \, dx = k\int_0^{\infty} \frac{1}{\sqrt{x^q+1}} \, dx.\]
\end{questionparts}\begin{questionparts}
\item \begin{align*}
&& I &= \int_0^{\infty} \frac{\sqrt{x}-1}{\sqrt{x(x^3+1)}} \, dx \\
x = u^{-1}, \d x = -u^{-2} \d u: && &= \int_{u=\infty}^{u = 0} \frac{u^{-1/2} - 1}{\sqrt{u^{-1}(u^{-3}+1)}} (-u^{-2}) \d u \\
&&&= \int_0^\infty \frac{u^{-1/2} -1}{\sqrt{1+u^3}} \d u \\
&&&= \int_0^\infty \frac{1-\sqrt{u}}{\sqrt{u(u^3+1)}} \d u \\
&&&= -I \\
\Rightarrow && I &= 0
\end{align*}
\item \begin{align*}
&& I &= \int_0^{\infty} \frac{1}{\sqrt{x^3+1}} \, dx \\
x = u^{-2}, \d x = -2u^{-3} \d u: && &= \int_{u = \infty}^{u = 0} \frac{1}{\sqrt{u^{-6}+1}} (-2u^{-3}) \d u \\
&&&= \int_0^\infty \frac{2}{\sqrt{1+u^6}} \d u \\
&&&= 2\int_0^{\infty} \frac{1}{\sqrt{x^6+1}} \, \d x
\end{align*}
\item \begin{align*}
&& I &= \int_0^{\infty} \frac{x^r - 1}{\sqrt{x^s(x^p+1)}} \, dx \\
x = u^{-1}, \d x = - u^{-2} \d t: &&&= \int_{u=\infty}^{u = 0} \frac{u^{-r}-1}{\sqrt{u^{-s}(u^{-p}+1)}} (- u^{-2}) \d u \\
&&&= \int_0^\infty \frac{u^{-r}-1}{\sqrt{u^{4-s}(u^{-p}+1)}} \d u \\
&&&= \int_0^\infty \frac{1-u^{r}}{\sqrt{u^{4-s+2r}(u^{-p}+1)}} \d u \\
&&&= \int_0^\infty \frac{1-u^{r}}{\sqrt{u^{4-s+2r-p}(1+u^p)}} \d u \\
\end{align*}
Therefore if $4-s+2r-p = s$ or $r = \frac{p}2+s-2$ we have $I = -I$, ie $I = 0$.
\item \begin{align*} && I &= \int_0^{\infty} \frac{1}{\sqrt{x^p+1}} \, dx \\
x = u^{-t}, \d x = -t u^{-t-1}:&&&= \int_{u = -\infty}^{u = 0} \frac{1}{\sqrt{u^{-pt}+1}} (-t u^{-t-1}) \d u \\
&&&= t \int_0^\infty \frac1{\sqrt{u^{-pt+2t+2}+u^{2t+2}}} \d u
\end{align*}
Therefore if $\begin{cases} 2t+2 &= q \\ 2t+2 -pt &= 0 \end{cases}$, ie $q = 2t+2, p = 2 + 2/t$ we will have found the $p, q$ desired for any $t$ (or $k$).
[Alternatively]
Let $\displaystyle I(p) =\int_0^{\infty} \frac{1}{\sqrt{x^p+1}} \, dx$, then clearly $I(p)$ is decreasing and as $p \to \infty$ $I(p) \to 1$, so our integral can take any values on $(1, \infty)$ and so for any positive value we can find two values with a given ratio. In particular given $k$ and $p$ we can find a suitable $q$.
\end{questionparts}
This was the second most popular question after Question 1 and was attempted by the vast majority of candidates. In general, solutions were very good, particularly for the first two parts. Parts (i) and (ii) were answered well, with candidates generally showing a good level of proficiency with completing the given substitutions. In part (i) a small number of candidates did not give enough detail in their method when dealing with the limits of the integral following the substitution. Most recognised that the substitution could be used to show that I = −I and produced clear explanations of this. Solutions to part (ii) were often fully correct. Those candidates who were able to identify the correct substitution were often successful in solving part (iii), although in some cases errors were made with the indices when simplifying the expression. Some candidates attempted substitutions which did not allow them to make any significant progress on solving this part of the question. Part (iv) was generally answered more successfully than part (iii) with most candidates able to identify the correct substitution to be made. Some candidates started with a more general substitution, from which the form that was needed was deduced. The substitution was again completed successfully by most candidates who reached this part, and the most complete responses noted that the change to the limits would be valid for any of the appropriate values for k. Having completed the substitution, many were able to identify a possible pair of values for p and q. Those who tried to argue that such a pair must exist often did not explain their reasoning clearly enough.