Find $\int_1^a \sqrt[5]{x^5-1}\ dx + \int_0^b \sqrt[5]{x^5+1}\ dx$, where $a^5-b^5 = 1$ So I am preparing to go to this olympiad, the Latvian Sophomore's dream calculus olympiad. I received previous years problems and the toughest problem in the definite integral section was this
$$\text{Find } \int_1^a \sqrt[5]{x^5-1}\ dx  + \int_0^b \sqrt[5]{x^5+1}\ dx$$
$$\text{where }\  a^5-b^5 = 1$$
I tried substituting the whole root sign in the respective integrals but that led to nowhere. I don't see how trigonometric substitution could be used, dummy variables or the DI method. I am really at a loss here.
Any ideas?
I added a picture of all the problems.

 A: $$a^5-b^5=1\Rightarrow  a=\sqrt[5]{b^5+1}$$
$$\text{let } \sqrt[5]{x^5-1}=t\Rightarrow x=\sqrt[5]{t^5+1}\Rightarrow dx=(\sqrt[5]{t^5+1})'dt$$
$$\Rightarrow \color{blue}{\int_1^{\sqrt[5]{b^5+1}}\sqrt[5]{x^5-1}\,dx}=\int_0^b t (\sqrt[5]{t^5+1})'dt\overset{IBP}=t \sqrt[5]{t^5+1}\bigg|_0^b-\color{red}{{\int_0^b \sqrt[5]{t^5+1}\,dt}}$$
$$\overset{\color{red}{t=x}}\Rightarrow \color{blue}{\int_1^a\sqrt[5]{x^5-1}\,dx}+\color{red}{\int_0^b \sqrt[5]{x^5+1}\,dx}=b\sqrt[5]{b^5+1}=ab$$
A: Lemma. Let $[c,d]\subset\mathbb R$ be an interval and $$f:[c,d]\to f([c,d])$$ be a bijective function with integrable derivative $f'$. Let $f^{-1}$ be the inverse of $f$. Then $$\int_c^d f(x)\,\mathrm dx+\int_{f(c)}^{f(d)} f^{-1}(x)\,\mathrm dx=df(d)-cf(c).$$
Proof. Since $f$ is continuous, $f([c,d])$ also is an interval. Hence by substitution and integration by parts, $$\int_{f(c)}^{f(d)} f^{-1}(x)\,\mathrm dx=\int_c^d x f'(x)\,\mathrm dx=\big[xf(x)\big]_{c}^d-\int_c^d f(x)\,\mathrm dx=df(d)-cf(c)-\int_c^d f(x)\,\mathrm dx.$$
This completes the proof. $\square$
In our particular case, we have $f:[1,a]\to[0,b]$ with $f(x)=\sqrt[5]{x^5-1}$ which indeed suffices all conditions of the lemma.
So $$\int_1^a \sqrt[5]{x^5-1}\,\mathrm dx \ +\ \int_0^b \sqrt[5]{x^5+1}\,\mathrm dx=af(a)-f(1)=a\sqrt[5]{a^5-1}-0=ab.$$
