The equation of a curve is $y=f(x)$. The tangents at $(1, f(1)),(2, f(2))$ and $(3, f(3))$ make angles $\frac{\pi}{6}, \frac{\pi}{3}$ and $\frac{\pi}{4}$ respectively with the positive direction of the x-axis. Then, the value of $\int\limits_2^3 f'(x) f''(x) d x+\int\limits_1^3 f''(x) d x$ is equal to

$-\frac{1}{\sqrt{3}}$

We have,

$f'(1) =\tan \frac{\pi}{6}=\frac{1}{\sqrt{3}}, f'(2)=\tan \frac{\pi}{3}=\sqrt{3}$ and $f'(3)=\tan \frac{\pi}{4}=1$

∴  $\int\limits_2^3 f'(x) f''(x) d x+\int\limits_1^3 f''(x) d x$

$=\int\limits_2^3 f'(x) d\left(f'(x)\right)+\int\limits_1^3 d\left(f'(x)\right)$

$=\left[\frac{\left\{f'(x)\right\}^2}{2}\right]_1^3+\left[f'(x)\right]_1^3$

$=\frac{1}{2}\left[\left\{f'(3)\right\}^2-\left\{f'(2)\right\}^2\right]+\left[f'(3)-f'(1)\right]$

$=\frac{1}{2}(1-3)+\left(1-\frac{1}{\sqrt{3}}\right)=-\frac{1}{\sqrt{3}}$