Challenging Integrals for High School Students I am now in my last year of high school. We have covered all the techniques useful for indefinite integration that are included in our Maths and Further Maths courses. This includes:

*

*Integration by parts, inspection, substitution, partial fraction decomposition


*Integration of regular and inverse trigonometric, regular and inverse hyperbolic, exponential, logarithmic, polynomial functions
I would like to have some challenging integrals to attack that are possible for me to solve at my current level of knowledge. By challenging, I mean integrals similar to the ones in this document. They were generally enjoyable and very satisfying to solve. If you have an integral that you think I could do that is more challenging than those in the aforementioned link, so much the better.
Thank you for your suggestions.
 A: These problems are not challenging but still good to do them

*

*$\int\exp(x)\bigg(\frac{1+x\ln x}{x}\bigg)\mathrm dx$


*$\int \sin(101x) \sin^{99}(x)\mathrm dx$


*$\int \sqrt{x-\sqrt{x^2-4}} \ \ \mathrm dx$
Play with these they are high school level.
A: In this question, $a$ is a positive constant.
(i)        Express $\cosh a$ in terms of exponentials.
             By using partial fractions, prove that $$\int_0^1 \frac{1}{x^2 + 2x\cosh a + 1} \mathrm dx = \frac{a}{2 \sinh a} .$$
(ii)       Find, expressing your answers in terms of hyperbolic functions,
$$\int_0^\infty \frac{1}{x^2 + 2x \sinh a -1} \mathrm dx$$
and
$$\int_0^ \infty \frac{1}{x^4 +2x^2 \cosh a + 1} \mathrm dx.$$

Find the area of the region between the curve $y = \frac {\ln x}{x} $
and the $x$-axis, for $1 \leq x  \leq a$. What happens to this area as $a$ tends to infinity?
Find the volume of the solid obtained when the region between the curve $y = \frac{\ln x}{x}$ and the $x$-axis, for $1 \leq x \leq a$, is rotated through $2π$ radians about the $x$-axis. What happens to this volume as $a$ tends to infinity?

For $n = 1, 2, 3, ..., $ let
$$I_n = \int_0^1 \frac{t^{n-1}}{(t+1)^n} \mathrm dt. $$
By considering the greatest value taken by $ \frac{t}{t+1} $ for $0 \leq  t \leq 1$ show that $I_n+1 < \frac{1}{2}I_n .$
Show also that $I_{n+1} = -\frac{1}{n2^n-1}.$
Deduce that $I_n < \frac{1}{n2^n-1}$
Prove that
$$ \ln 2 = \sum_{r=1}^n \frac{1}{r 2^r} + I_{n+1}$$
and hence how that $\frac{2}{3} < \ln 2 < \frac{17}{24}.$

The above questions are sourced from this website. All credits go to them.
If you need any more questions, please just drop a comment.
A: Not quite an integral, but a differential equation. This one definitely requires some outside of the box thinking!

Consider the solution $y(x)$ of the differential equation
$$
\frac{dy}{dx}=\sqrt{1+y^{2020}}
$$
determined by the initial data $y(0)=0$.
Prove that for all $n \in \mathbb{N}$, $y^{(n)}(0)$ is an integer. Then find the first non-zero value of $y^{(n)}(0)$ with $n>1$.
A: Since you asked for integrals, here is an integral! This one I found quite challenging, but perhaps you'll find a more elegant way of solving it than I did:
$$
\int \frac{dx}{(9-36x^2)^{3/2}} \, .
$$
NB this definitely can be solved using the methods outlined in your post.
A: Here are a couple of calculus questions concerning hyperbolic functions:

*

*Just as $\sin$ can be defined as the unique function $f:\mathbb{R} \mapsto \mathbb{R}$ satisfying
\begin{align}
f''(x) &= \color{blue}{-}f(x) \\
f'(x) &= 1 \\
f(x) &= 0 \, ,
\end{align}
$\sinh$ can be defined as the unique function $f:\mathbb{R} \mapsto \mathbb{R}$ satisfying
\begin{align}
f''(x) &= \color{red}{+}f(x) \\
f'(x) &= 1 \\
f(x) &= 0 \, .
\end{align}
Derive the exponential form of hyperbolic sine by solving the above equation.

*The classical way to define $\sinh$ is via the 'unit hyperbola':

If the region OPR has an area of $t/2$, then the $x$- and $y$-coordinates are, respectively, $\cosh t$ and $\sinh t$. (If you're interested in the details of how hyperbolic functions can be defined using a hyperbola, then see here.) If however you define $\sinh$ and $\cosh$ using exponentials, then it is possible to prove that the shaded area is equal to $t/2$, given that $OQ = \cosh t$ and $PQ = \sinh t$. Prove this using integration. You may need to use hyperbolic identities to simplify your answer.

*The graph of $\cosh$ has a curious property. For any interval $[a,b]$, the area under the graph is equal to the length of the arc connecting the points $(a,\cosh a)$ and $(b,\cosh b)$. Prove this using the formula
$$
\text{arc length} = \int_{a}^{b}\sqrt{1+\left(f'(x)\right)^2} \, dx \, ,
$$
where $y=f(x)$ is the graph of the function in question. (Here, $f=\cosh$.) And if you're interested in seeing why the arc length formula works, then see here.

A: This one's really nice.
$$\int\frac{1}{x^7-x}\text{ }dx$$
There's a clever trick that can save you from a tedious partial fraction decomposition. Can you find it?
A: To add yet another answer to your question, here are two integrals which are challenging, but their solutions are not too messy:
\begin{align}
&\int \sin(\log_2(x)) \, dx \, , \\[5pt]
&\int \frac{dx}{\sqrt{e^{2x}+1}} \, .
\end{align}
