(c) : $f(x)=\int \frac{x^2-3 x+2}{x^4+1} d x$ $f(x)=\int \frac{(x-2)(x-1)}{x^4+1} d x$, then $f^{\prime}(x)=\frac{(x-2)(x-1)}{x^4+1}$ For critical points, $f^{\prime}(x)=0$ $ \Rightarrow(x-2)(x-1)=0 \quad\left[\because x^4+1>0 \forall x \in R \right]$ $\Rightarrow x=2,1$ are critical points. So, $f^{\prime \prime}(x)=\frac{(2 x-3)\left(x^4+1\right)-4 x^3\left(x^2-3 x+2\right)}{\left(x^4+1\right)^2}$ $f^{\prime \prime}(1)=\frac{-2-4 \times 0}{2^2}=\frac{-1}{2}<0$ $\Rightarrow x=1$ is the point of local maxima. $f^{\prime \prime}(2)=\frac{17-32 \times 0}{(17)^2}=\frac{1}{17}>0$ $\Rightarrow x=2$ is the point of local minima. $\therefore f(x)$ is decreasing in $(1,2)$
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