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. 
 A: 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:
$$
\int f'(ax+b)\, dx = \frac{1}{a}f(ax+b)+C.
$$
A: If differentiate back we get $\frac{d}{dx}(\frac{(1+x^2)^5}{10x}+C)=\frac{10x(5(2x)(1+x^2)^4)-10(1+x^2)^5}{100x^2}\ne (1+x^2)^4$.
You were trying to apply the rule  $\frac{d}{dx}(f(x))^{n+1})=(n+1)f'(x)f(x)^n$, but one cannot do the following:
$\int f(x)^ndx=\int \frac{1}{f'(x)}f'(x)f(x)^ndx=\frac{1}{f'(x)}\int 
f'(x)f(x)^ndx$.
The final equality is wrong, we cannot bring "$f'(x)$" outside of the integral, and this is where your error came in.
A: you can use the rule after expanding
$\int (1+x^2)^4dx=\int 
(1+2x^2+x^4)^2=\int(1+4x^4+x^8+2x^4+4x^2+4x^6)dx\\=x+4x^5/5+x^9/9+2x^5/5+4x^3/3+4x^6/6+C$
A: You are integrating with respect to $x$, not $(1+x^2)$. It is why it doesn't work the way you think. U-substitution makes this possible. So in a way it is kind of what you expect. Just a little change of variables. 
