This is from my previous thread:
$\int(1+x^2)^4\mathrm dx$ $\ne$ $\frac{(1+x^2)^{5}}{5(2x)}+C$?
because differentiating back gives $\frac{d}{dx}(\frac{(1+x^2)^5}{10x}+C)=\frac{10x(5(2x)(1+x^2)^4)-10(1+x^2)^5}{100x^2}\ne (1+x^2)^4$.
The best answer was that I was differentiating with respect to $x$ and $not$ $(1+x^2)$. I still don't fully understand why this is so, and I am looking for some more advice and intuition on why the inner function must be linear to use the power rule.
A big thanks to all those that answered in the last thread.
It looks like you wanted to use the power rule on $(1 + x^2)$, for example, putting $u = 1+x^2$, then integrating with respect to $u$ requires that we first find $du = 2x \,dx\iff dx = \dfrac{du}{2x}$, in order to replace $dx$ in the original integral. This is where the fact that $1+x^2$ is not linear begins to create problems.
(Hmmm, looks like a problem, since if we replace dx with its equivalent, we still have a factor of $x$.) Okay, so let's try to express $x$ in terms of $u$: $$u = 1+x^2 \iff x^2 = u-1 \implies x = \pm \sqrt{u - 1}$$ That gives us $$dx = \dfrac {du}{2(\pm \sqrt{u - 1})}$$
(Hmmm, looks like this is getting complicated.)
So we have that $$\int (1 + x^2)^4 \,dx = \int u^4 \cdot \frac{du}{\pm 2\sqrt{u - 1}} = \pm \int \dfrac{u^4}{2\sqrt{u - 1}}\,du$$
But this messes up our hopes to use the power rule! We can't ignore the denominator and simply apply the power rule to the numerator.
So a better plan is to go back to the original integral, using trigonometric substitution by setting $u = \tan \theta \implies du = \sec^2 \theta.$