$I_1=\int \frac{x^3}{{\left(1+x^2\right)}^3}dx = f\left(x\right)+k$ 

$x=\tan \theta$

$\Rightarrow dx={\sec }^2\theta d\theta$

$I_1=\int \frac{{\tan }^2\theta {\sec }^2\theta d\theta }{{\left(1+{\tan }^2\theta \right)}^3}$

$I_1=\int \frac{{\tan }^3\theta {\sec }^2\theta d\theta }{{\sec }^6\theta }$

$I_1=\int {\sin }^3\theta \cos \theta d\theta$

Let $\sin \theta =p$

$\Rightarrow \cos \theta d\theta =dp$

$I_1={\int }_{}^{}{p}^{3}dp=\frac{p^4}{4}+k$

$\Rightarrow I_1=\frac{{\sin }^4\theta }{4}+k$

$\Rightarrow I_1=\frac{{\sin }^4\left[{\tan }^{-1}x\right]}{4}+k$

Let ${\tan }^{-1}x=\alpha \Rightarrow x=\tan \alpha$

$\Rightarrow \sin \alpha =\frac{x}{\sqrt{1+x^2}}$

$\Rightarrow \frac{{\sin }^4\alpha }{4}=\frac{x^4}{4\left(1+x^2\right)}+k=f\left(x\right)+k$

Now $f\left(x\right)=\frac{x^4}{4\left(1+x^2\right)}$

$I_1={\int }_{}^{}\frac{x^3}{{\left(1+x^2\right)}^3}dx =g\left(x\right)+c$

Let $1+x^2=z\Rightarrow x^2=z-1$

$\Rightarrow xdx=\frac{1}{2}dz$

$I_1=\frac{1}{2}\int \frac{z-1}{z^3}dz$

$\Rightarrow I_1=\frac{1}{2}\int \left(\frac{1}{z^2}-\frac{1}{z^3}\right)dz$

$I_1=\frac{1}{2}\left[\frac{-1}{z}+\frac{1}{2z^2}\right]$

$\Rightarrow I_1=\frac{1}{2}\left[\frac{-1}{1+x^2}+\frac{1}{2{\left(1+x^2\right)}^2}\right]$

$\Rightarrow I_1=\left[\frac{-2x^2-1}{4{\left(x^2+1\right)}^2}\right]+c=\left[g\left(x\right)+c\right]$

$\Rightarrow g\left(x\right)=\left[\frac{-2x^2-1}{4{\left(x^2+1\right)}^2}\right]$

Now $f\left(x\right)-g\left(x\right)+k-c=$$\frac{x^4}{4\left(1+x^2\right)}-\left[\frac{-2x^2-1}{4{\left(x^2+1\right)}^2}\right]$

$=\frac{x^4+2x^2+1}{4{\left(x^2+1\right)}^2}+k-c$

$=\frac{1}{4}+k-c$

Hence, it is a constant.