Making integration problems such that $\int{}f^{-1}(x)\text{ d}x$ is difficult and $\int{}f(x)\text{ d}x$ is easy.

Question

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 $\int{f(x)\text{ d}x}=\color{green}{\text{EASY}}$

Answer 1

As pointed out in the comments, any such integral $\int f^{-1}(x) \,dx$ can be handled by applying integration by parts---essentially the identity mentioned at the beginning of the post---with $$u = x \qquad v = f^{-1}(x) .$$ So, one way to make a "hard" example is to construct a function whose inverse is not immediately apparent. For example, if we take $$f(x) = x^3 + x + 1,$$ Cardano's Formula gives that $$f^{-1}(x) = \frac{\sqrt[3]{-108 + 108 x + 12 \sqrt{81 x^2 - 162 x + 93}}}{6} - \frac{2}{\sqrt[3]{-108 + 108 x + 12 \sqrt{81 x^2 - 162 x + 93}}} .$$ This latter function is perhaps not so easy to invert, but we can compute $$F(x) = \frac{1}{4} x^4 + \frac{1}{2} x^2 + x + C$$ and so \begin{multline} \small \int \left(\frac{\sqrt[3]{-108 + 108 x + 12 \sqrt{81 x^2 - 162 x + 93}}}{6} - \frac{2}{\sqrt[3]{-108 + 108 x + 12 \sqrt{81 x^2 - 162 x + 93}}}\right) \,dx \\ \small = x \left(\frac{\sqrt[3]{-108 + 108 x + 12 \sqrt{81 x^2 - 162 x + 93}}}{6} - \frac{2}{\sqrt[3]{-108 + 108 x + 12 \sqrt{81 x^2 - 162 x + 93}}}\right) \\ \small - \left[\frac{1}{4}\left(\frac{\sqrt[3]{-108 + 108 x + 12 \sqrt{81 x^2 - 162 x + 93}}}{6} - \frac{2}{\sqrt[3]{-108 + 108 x + 12 \sqrt{81 x^2 - 162 x + 93}}}\right)^4 \\ \small + \frac{1}{2} \left(\frac{\sqrt[3]{-108 + 108 x + 12 \sqrt{81 x^2 - 162 x + 93}}}{6} - \frac{2}{\sqrt[3]{-108 + 108 x + 12 \sqrt{81 x^2 - 162 x + 93}}}\right)^2 \\ \small + \left(\frac{\sqrt[3]{-108 + 108 x + 12 \sqrt{81 x^2 - 162 x + 93}}}{6} - \frac{2}{\sqrt[3]{-108 + 108 x + 12 \sqrt{81 x^2 - 162 x + 93}}}\right)\right] + C\end{multline}

Answer 2

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*}