Question 0:$$\int x^{42} + 2x + 1 \,dx$$

Answer: $$\frac{x^{43}}{43} + x^2 + x + C$$

This was the demo question we gave to show how to submit answers.

Question 1: $$\int \left(\frac{x}{2023}+1\right)^{2023}\,dx$$

Answer: $$\frac{2023}{2024}\left(\frac{x}{2023}+1\right)^{2024} + C$$

Substitute $u=\frac{x}{2023}+1$.

Question 2: $$\int 5x^2 + \sin(2x) + 3\,dx$$

Answer: $$\frac{5x^3}{3} - \frac{\cos(2x)}{2} + 3x + C$$

Question 3: $$\int \left(e^x\right)^2\,dx$$

Answer: $$\frac{e^{2x}}{2} + C$$

Recall that $\left(e^x\right)^2=e^{2x}$. This took much longer to solve than expected; the nerves of solving problems infront of a crowd don't lend themselves well to integation.

Question 4: $$\int \left(x-\sqrt[3]{2}\right)\left(x^2 + \sqrt[3]{2}x + 2^\frac{2}{3}\right)\,dx$$

Answer: $$\frac{x^4}{4} - 2x + C$$

This one is much easier if you recall the difference of cubes formula, so: \begin{align*}&\phantom{{}={}}\left(x-\sqrt[3]{2}\right)\left(x^2 + \sqrt[3]{2}x + 2^\frac{2}{3}\right)\\&=x^3-\left(\sqrt[3]{2}\right)^3\\ &=x^3-2\end{align*}

Question 5: $$\int 9x\sin\left(2x^2\right)\,dx$$

Answer: $$-\frac{9}{4}\cos\left(2x^2\right) + C$$

Substitute $u=2x^2$.

Question 6: $$\int_0^1 \frac{8}{3}x^4 + \frac{4}{5}x^3 + \frac{7}{3}x^2 - \frac{3}{5}x -\frac{11}{180} \, dx$$

Answer: $$\frac{23}{20}$$

I didn't make this one (it's the work of Ben Varley), but it worked well in the intial rounds. Got quite a few laughs and half-hearted sighs from the audience.

Question 8: $$\int -7x^2\left(x^3+7\right)^7 \,dx$$

Answer: $$-\frac{7}{24}\left(x^3+7\right)^8 + C$$

Substitute $u=x^3+7$.

Question 9: $$\int_0^{2\ln 2} \sqrt{e^x}\,dx$$

Answer: $$2$$

For this one, recall that $\sqrt{e^x}=e^\frac{x}{2}$. The bounds of integration are often forgotten when people are in the heat of the moment.

Question 10: $$\int_0^{1} \cos(\pi x) + \sin(\pi x) + \cos(\pi x)\sin(\pi x)\,dx$$

Answer: $$\frac{2}{\pi}$$

A very common trick is to be aware odd and even functions. Here the first and third terms vanish to 0, making it a bit easier.

Question 11: $$\int \cos^2 x - \sin^2 x \,dx$$

Answer: $$\cos x \sin x + C = \frac{\sin(2x)}{2}+C$$

Another one that Ben Varley wrote (it's quite beautiful). Using $\sin^2 x=1-\cos^2 x$ and $\cos^2 x = \frac{\cos(2x)+1}{2}$ we have: \begin{align*}\int \cos^2 x - \sin^2 x \,dx &= \int 2\cos^2(x) - 1 \,dx\\&= \int \cos(2x) \,dx\end{align} Substituting $u=2x$ then leads to the solution.

Question 12: $$\int_{-\frac{1}{2}\ln(e^2)}^{1}\operatorname{sign}(x) \cosh x\, dx$$

Answer: $$0$$

The product of $\operatorname{sign}$ (which is odd) and $\cosh$ (which is even) is odd, so the symmetry of the bounds provides an easy solve.

Question 13: $$\int 2023^x + x^{2023} \, dx$$

Answer: $$\frac{x^{2024}}{2024} + \frac{2023^x}{\ln 2023} + C$$

The first term is easier to solve once you remeber that $2023^x=e^{2023\ln x}$ (or if you remeber that $\int a^x\,dx = \frac{a^x}{\ln a}+C$).

Question 14: $$\int \left(2x^2+1\right)\cos x \,dx$$

Answer: $$\left(2x^3-3\right)\sin x + 4x\cos x + C$$

The first one needing by parts. If this is known, it's not too tricky: \begin{align*}&\phantom{{}={}}\int \left(2x^2+1\right)\cos x \,dx\\ &= (2x^2+1)\sin x - \int4x\sin(x)\,dx\\ &= (2x^2+1)\sin x + 4x\cos x - \int 4\cos x\,dx\\ &=\left(2x^3-3\right)\sin x + 4x\cos x + C\end{align*}

Question 15: $$\int_{-\pi}^\pi\int_{-\pi}^\pi x^2 + 3xy^2 + \sin y \,dx\,dy$$

Answer: $$\frac{4\pi^4}{3}$$

The first double integral. Notice that the second and third terms vanish over the $x$ and $y$ domains respectively, and that we can split up the $x$ and $y$ components of the first term, so: \begin{align*}\int_{-\pi}^\pi\int_{-\pi}^\pi x^2\, dx\,dy &= 2\pi\int_{-\pi}^\pi x^2\, dx\\ &= 2\pi\times \frac{2\pi^3}{3}\\ &= \frac{4\pi^4}{3}\end{align*}

Question 16: $$\int \ln(x-1) \,dx$$

Answer: $$(x-1)\ln(x-1) - x + C$$

This one might have been a bit too easy, as one of the contestants at the whiteboard knew the integral of $\ln$ (if you don't know it, consider doing by parts on $1\times\ln(x)$).

Question 17: $$\int e^x\sqrt{1+e^x} \,dx$$

Answer: $$\frac{2}{3}\left(1+e^x\right)^\frac{3}{2} + C$$

Substituting $u=1+e^x$ provides the solution quite quickly.

Question 18: $$\int \frac{d{\scriptstyle\int}}{\sqrt{1 + \!{\scriptstyle\int}}}$$

Answer: $$2\sqrt{{\scriptstyle\int} + 1} + x$$

Things like this always get giggles and groans from the audience. Otherwise a simple integral given the substitution $u={\scriptstyle\int}+1$.

Question 19: $$\iint_{x^2+y^2\leq \pi^2} 4x^2 + 8xy + 4y^2 \,dx\,dy$$

Answer: $$2\pi^5$$

This one isn't too bad if you convert to polar coordinates. Notice that $4x^2 +8xy +4y^2 = 4(x+y)^2 = 4r^2$ so: \begin{align*}&\phantom{{}={}}\iint_{x^2+y^2\leq \pi^2} 4x^2 + 8xy + 4y^2 \,dx\,dy\\ &= 2\pi\int_0^{\pi}4r^2\times r\,dr\\ &= 2\pi^5\end{align*}

Question 20: $$\int e^x\sin(\pi x) \,dx$$

Answer: $$\frac{e^x(\sin(\pi x)-\pi\cos(\pi x))}{1 + \pi^2} + C$$

This was a tricky one (that by parts eventually solves). It was well placed in the later rounds and almost didn't get solved in the 3 minutes given.

If we let $I$ be the original integral, we have: \begin{align*}I &= \int e^x\sin(\pi x) \,dx\\ &= -\frac{e^x\cos(\pi x)}{\pi} + \int\frac{e^x\sin(\pi x)}{\pi^2}\, dx\end{align*} Therefore $$(\pi^2+1)I = e^x(\sin(\pi x) - \pi\cos(\pi x))$$ and so by rearranging for $I$ we get the desired result.

Question 21: $$\int \frac{1-x}{1-\sqrt{x}} \,dx$$

Answer: $$\frac{2x^{\frac{3}{2}}}{3} + x + C$$

This one isn't so bad once you realise that: \begin{align*}\int \frac{1-x}{1-\sqrt{x}} \,dx &= \int \frac{\left(1-\sqrt{x}\right)\left(1+\sqrt{x}\right)}{1-\sqrt{x}} \,dx\\ &= \int 1+\sqrt{x}\,dx\end{align*}

Question 22: $$\int \tan^2x \,dx$$

Answer: $$\tan x - x + C$$

This one is the first one that requires some more knowledge of trig to complete. First, note that $\tan^2 x = \sec^2x - 1$. If you know the integral of $\sec^2x$ (namely $\tan x$), this then becomes easy.

Question 23: $$\int_0^1 x^3e^{\left(x^2\right)} \,dx$$

Answer: $$\frac{1}{2}$$

Start by substituting $u=x^2$, getting the equal integral of $$\int_0^1\frac{ue^u}{2}\,du.$$ Then using parts, this simplifies to $$\left[\frac{ue^u}{2}\right]_{u=0}^{u=1}-\int_0^1 \frac{e^u}{2}\,du,$$ which resolves down to $$\frac{e}{2}-0-\frac{e}{2}+\frac{1}{2}=\frac{1}{2}.$$

Question 24: $$\int \frac{1}{1 + e^x} \, dx$$

Answer: $$x - \ln(1 + e^x) + C}$$

After substituting $u=e^x$, we get $$\int\frac{du}{u(u+1)}=\int\frac{1}{u}-\frac{1}{u+1}\,du.$$

This is the first question using partial fractions. I tried to avoid it as much as possible, as it's a bit of a time sink and not very interesting to watch. However, I made an exception in this case, as it's simple and quite late into the questions.

Question 25: $$\int \frac{1}{x\ln x} + \frac{\ln x}{x} \,dx$$

Answer: $$\frac{\left(\ln x\right)^2}{x} + \ln\left|\ln x\right| + C$$

Both terms can be simplified by the substitution $u=\ln x$, leaving the integral $$\int \frac{1}{u} + u\,du.$$

Question 26: $$\int \left(e^{\sqrt{x}}\right)^2\,dx$$

Answer: $$\frac{\left(2\sqrt{x}-1\right)e^{2\sqrt{x}}}{2} + C$$

Noticing that $\left(e^\sqrt{x}\right)^2=e^{2\sqrt{x}}$, we can substitute $u=2\sqrt{x}$, getting the integral $$\int\frac{ue^u}{2}\,du.$$ Then by parts we get $$\frac{ue^u}{2}-\int\frac{e^u}{2}\,du = \frac{(u-1)e^u}{2}+C.$$

Question 27: $$\int x\sqrt{x-1} \,dx$$

Answer: $$\frac{2x(x-1)^\frac{3}{2}}{5} + \frac{4(x-1)^\frac{3}{2}}{15} + C$$

After substituting $u=x-1$, this simplifies to $$\int (u+1)\sqrt{u}\,du = \int u^\frac{3}{2}+\sqrt{u}\,du.$$

Question 28: $$\int \left(\ln x + 1\right)x^{2x} \,dx$$

Answer: $$\frac{x^{2x}}{2}+C$$

This was one of my favourites to come up with. The secret (and this one very much relies upon a secret) is knowing the derivative of $x^x$. As we have $$\frac{d}{dx}x^x = \frac{d}{dx}e^{x\ln x} = (1+\ln x)x^x$$. This leads to the realisation that $$\frac{d}{dx}x^{2x}=(1+\ln x)2x^{2x},$$ thus providing the answer.

Question 29: $$\int_{-\infty}^\infty x\cos\left(x^2\right)e^{-x^2} \,dx$$

Answer: $$0$$

Notice that $x$ is odd whereas $\cos\left(x^2\right)$ and $e^{-x^2}$ are even, so the integrand is odd and so vanishes over a symmetric domain.

Question 30: $$\int_{0}^\infty x\cos\left(x^2\right)e^{-x^2} \,dx$$

Answer: $$\frac{1}{4}$$

This one follows well from the previous one. Starting by substituting $u=x^2$, we get $$\int_0^\infty\frac{e^{-u}\cos u}{2}\,du,$$ which we will call $I$. Integration by parts will then reduce it down to: \begin{align*}I&=\left[\frac{e^{-u}\sin u}{2}\right]_{u=0}^{u=\infty}+ \int_0^\infty\frac{e^{-u}\sin u}{2}\\ &= 0 - 0 - \left[\frac{e^{-u}\cos u}{2}\right]_{u=0}^{u=\infty}\\ &\phantom{{}={}}\qquad - \int_0^\infty\frac{e^{-u}\cos u}{2}\,du\\ &= -0 + \frac{1}{2} - I\end{align*} This means that our original integral $I$ is $\frac{\pi}{4}$.

This ended up as the final integral this year. I think it was a good finale, with both contestants making some mistakes before one got the answer in the nick of time.

Question 31: $$\int_0^3 \left(9x^2 + 3\right)\operatorname{sign}(\sin(\pi x)) \,dx$$

Answer: $$42$$

This one needs a bit of care, by breaking it up into 3 parts. First, we have: $$\int 9x^2+3\,dx = 3x^3+3x$$ From $0$ to $1$, $\sin(\pi x)$ is positive, so this part of the integral is $3+3-0=6$. From $1$ to $2$, $\sin(\pi x)$ is negative, so we need to subtract $3\times 8+3\times 2 - 6 = 24$. Finally, from $2$ to $3$, $\sin(\pi x)$ is positive, so we will add on $3\times 27+3\times 3 - \left(3\times 8+3\times 2) = 60$. Together, the total integral is $6-24+60=42$.

Question 32: $$\int_0^{\frac{\pi}{2}} \frac{8\sin x}{\sin x + \cos x\right)}\,dx$$

Answer: $$2\pi$$

I think this is my favourite trick. I used it a lot last year. Notice that if we substitute $u=\frac{\pi}{2}-x$ then $\sin x=\cos u$, we get the integral: $$\int_{0}^{\frac{\pi}{2}}\frac{8\sin x}{\sin x + \cos x\right)}\,dx$$ Adding this to the original integral, we get $$\int_0^{\frac{\pi}{2}}8\,dx = 4\pi.$$ As this is twice the original integral, we must have that it is $2\pi$.

Question 33: $$\int \sin\sqrt{x} \,dx$$

Answer: $$2\sin\sqrt{x}-2\sqrt{x}\cos\sqrt{x}$$

Substituting $u=\sqrt{x}$ we get: $$\int 2u\sin u\,du$$ The rest then follows from integration by parts.

Question 34: $$\int_0^\infty \frac{x}{e^x-1} \,dx$$

Answer: $$\frac{\pi^2}{6}$$

First, we will rearrange to get: $$\int_0^\infty\frac{xe^{-x}}{1-e^{-x}}$$ As $\left|e^{-x}\right|<1$ for our domain, we can replace $\frac{1}{1-e^{-x}}$ with the geometric series $\sum_{k=0}^\infty\left(e^{-x}\right)^k$. As the integral is sufficiently nice, we can swap the summation and integral signs, to get: $$\sum_{k=0}^\infty\int_{0}^\infty xe^{-x}\left(e^{-x}\right)^k\,dx = \sum_{k=0}^\infty\int_{0}^\infty xe^{-(k+1)x}\,dx$$ Changing the bounds, we get that this is equal to $$\sum_{k=1}^\infty\int_{0}^\infty xe^{-kx}\,dx.$$ By doing the inner integral via integration by parts, we find that it equals $\frac{1}{k^2}$, so our original integral is equal to $\sum_{k=1}^\infty\frac{1}{k^2}$. This is the famous Basel sum, leading to such a nice answer.

This integral was inspired by a similar integral I saw solved online, I'm sure this is nothing new, but this technique really amazed me. I'll be trying in future to use it more.