Why is $\int(1+x^2)^4\mathrm {d}x$ not equal to $\frac{(1+x^2)^{5}}{5(2x)}+C$? This is a Rhetorical question since I already know that the answer is because the inner function: $$(1+x^2)$$ is not linear.
But what I want to know is a $why$ it must be linear. I have scoured the internet and maths books rigorously looking for an answer to this but no such luck. So if someone could kindly explain to me in simple english (if possible) why it must be linear and why $$\int(1+x^2)^4\mathrm dx$$ is not equal to $$\frac{(1+x^2)^{5}}{5(2x)}+C?$$
Any responses appreciated,
Many Thanks.
regards,
BLAZE. 

Do you remember the chain rule:
$$
(g \circ f)' = (g'\circ f)f'?
$$
This is the reason. If you need to compute
$$
\int f'(x+q)\, dx,
$$
the answer is 
$$
f(x+q)+C,
$$
since the derivative of $x+q$ is $1$. By the same token,
$$
\int f'(x^2)\, dx
$$
is not $$f(x^2)+C,$$ because the derivative of $x^2$ is $2x$, and hence $$f(x^2)'=2x f'(x^2),$$ a rather different function than $f'(x^2)$.
In general, linear transformations of the $x$ variable are rather harmless in integration theory, since you can always get rid of them by a simple substitution:
$$
\int f'(ax+b)\, dx = \frac{1}{a}f(ax+b)+C.
$$
