Integration by parts or substitution

Question

Let $ f:\mathbb{R}\rightarrow \mathbb{R}$ be a derivative function with continue derivative so that $f(a)=0$ and $f(b)=6$. Let $g(x)$ also be $g(x)=\int_{6}^{f(x)} \sqrt[4]{1+t^{2}} dt $

Then, the value of $ \int_{a}^{b} f'(x) g(x) dx$ is ...

I am having a hard time with this question... I tried using integration by parts using: $u=g(x)$ and $dv=f'(x)$ so that $$\int_{a}^{b} f'(x) g(x) dx = g(x) f(x) - \int_{a}^{b} f(x)g'(x)$$

Then $$\int_{a}^{b} f'(x) g(x) dx = \int_{6}^{f(x)}\sqrt[4]{1+t^{2}}dt f(x) - \int_{a}^{b} f(x)\sqrt[4]{1+t^{2}}dt$$

I'm not supposed to use Fubini's theorem, so I'm trying to solve it with integration by parts or with substitution

Answer 1

\begin{align*} \int_a^b f'(x) g(x)dx &= \left[f(x) g(x)\right]_a^b-\int_a^b f(x) g'(x)dx\\ & = f(b)g(b)-f(a)g(a)-\int_a^b f(x)f'(x)\sqrt[4]{1+f(x)^2}\\ &= f(b)\underbrace{g(b)}_{=0}-\underbrace{f(a)}_{=0}g(a) -\frac 12 \left[ \frac{(1+f(x)^2)^{5/4}}{5/4}\right]_a^b\\ & = -\frac 25 (37^{5/4}-1) \end{align*}