Making integration problems such that $\int{}f^{-1}(x)\text{ d}x$ is difficult and $\int{}f(x)\text{ d}x$ is easy. There is an amazing formula to integrate the inverse of a function:
$$\int{f^{-1}(x)\text{ d}x}=x f^{-1}(x)-F\bigg(f^{-1}(x)\bigg)+c, \text{where }F(x)=\int{f(x)\text{ d}x}$$
I know how to derive this formula. So if we have $\int{f^{-1}(x)\text{ d}x}$
Then starting with the substitution $u=f^{-1}(x)$
So $x=f(u)$ and $\text{d}x=f'(u)\text{ d}u$
Therefore, $\int{f^{-1}(x)\text{ d}x}=\int u f'(u) \text{ d}u$
Integrating by parts, $\int u f'(u) \text{ d}u=u f(u)-\int f(u) \text{ d}u=x f^{-1}(x)-F\bigg(f^{-1}(x)\bigg)+c$, where $F(x)=\int f(x) \text{ d}x$.

An example to use this is to find $\int \log(x) \text{ d}x$
We have $f^{-1}(x)=\log(x) \implies f(x)=e^x \implies F(x)=\int f(x) \text{ d}x=\int e^x \text{ d}x=e^x \implies F\bigg(f^{-1}(x)\bigg)=e^{\log(x)}=x$
So $\int \log(x) \text{ d}x=x \log(x)-x+c$

Another example is $\int \tanh^{-1}(x) \text{ d}x$
We have $f^{-1}(x)=\tanh^{-1}(x) \implies f(x)=\tanh(x) \implies F(x)=\int f(x) \text{ d}x=\int \tanh(x) \text{ d}x = \log\big|\cosh(x)\big|$
$\implies F\bigg(f^{-1}(x)\bigg)=\log\bigg|\cosh\big(\tanh^{-1}(x)\big)\bigg|$
So $\int \tanh^{-1}(x) \text{ d}x=x\tanh^{-1}(x)-\log\bigg|\cosh\big(\tanh^{-1}(x)\big)\bigg|+c$

The previous two examples are also easy without this amazing formula; the first example can be determined directly by integrating by parts (well-known classical example of integrating by parts), and the second example we can replace $\tanh^{-1}(x)$ with its equivalent expression:
$\frac{1}{2}\log\bigg(\frac{1+x}{1-x}\bigg)=\frac{1}{2}\bigg(\log(1+x)-\log(1-x)\bigg)$, then integrate by parts.

However, I am looking for integrals that cannot be done easily without the formula mentioned above (possibly can be done by a long and tedious way). For instance,
$$\int \sqrt{\frac{1}{x}-1}\text{ d}x$$
Let $f^{-1}(x)=\sqrt{\frac{1}{x}-1}$
Now using the usual way to find the inverse of a function (steps are omitted), we get $f(x)=\frac{1}{x^2+1}$
$F(x)=\int f(x) \text{ d}x= \tan^{-1}(x)$, so $F\bigg(f^{-1}(x)\bigg)=\tan^{-1}\bigg(\sqrt{\frac{1}{x}-1}\bigg)$
So $\int \sqrt{\frac{1}{x}-1}\text{ d}x=x \sqrt{\frac{1}{x}-1} - \tan^{-1}\bigg(\sqrt{\frac{1}{x}-1}\bigg)+c$

I do not want the integrand to be logarithm or inverse trigonometric function or inverse hyperbolic function, because they are easy in the first place. However, I need to make integral problems similar to the last example mentioned above.
So $\int{f^{-1}(x)\text{ d}x}=\color{red}{\text{DIFFICULT}}$ but $\int{f(x)\text{ d}x}=\color{green}{\text{EASY}}$

Another example:
Take $$f(x) = -\frac{1}{x \sqrt{1 - x^2}} , \qquad x \in \left(0, \frac{1}{\sqrt{2}}\right)$$
(we've restricted the domain of $f$ so that it has an inverse). Then,
$$f^{-1}(x) = \frac{1}{\sqrt{2}} \frac{\sqrt{x^2 + x \sqrt{x^2 - 4}}}{x} .$$ Then, we can transform the integral
$$\int f^{-1}(x)\,dx = \frac{1}{\sqrt{2}} \int \frac{\sqrt{x^2 + x \sqrt{x^2 - 4}}}{x} \,dx$$
via the substitution $u = x^2, du = 2 x\,dx$ (dropping the resulting overall constant) to
$$\int \frac{\sqrt{u + \sqrt{u^2 - 4 u}}}{u} \,du .$$ Since $f(x) = \frac{d}{dx} \operatorname{arsech} x$, we find that
\begin{align*}\int \frac{\sqrt{u + \sqrt{u^2 - 4 u}}}{u} \,du &= 2 \sqrt{x^2 + x \sqrt{x^2 - 4}} - 2 \sqrt{2} \operatorname{arsech} \left(\frac{1}{\sqrt{2}} \frac{\sqrt{x^2 + x \sqrt{x^2 - 4}}}{x}\right) + C\\
&= 2 \sqrt{u + \sqrt{u^2 - 4 u}} - 2 \sqrt{2} \operatorname{arsech} \sqrt{\frac{u + \sqrt{u^2 - 4 u}}{2 u}} + C\end{align*}
