Integration as the Reverse of Differentiation
Integration undoes differentiation: raise the power by one and divide. The one thing differentiation destroys is the constant, which is why every answer ends in + C.
Integration reverses differentiation
Differentiation takes a function and gives you its gradient function. Integration goes the other way: given the gradient function, it recovers the function it came from. The two operations undo each other, which is the content of the Fundamental Theorem of Calculus.
We write $\displaystyle\int f(x)\,\mathrm{d}x$ for "the function whose derivative is $f(x)$". The $\mathrm{d}x$ says the variable is $x$. In practice, integrating means asking yourself one question: what would I differentiate to get this?
Differentiating $x^2$ gives $2x$. But so does differentiating $x^2 + 7$, or $x^2 - 3$, because constants differentiate to zero. So when you run backwards from $2x$, there is no way to tell which constant was there to begin with. We cover every possibility by adding an unknown constant of integration: $\displaystyle\int 2x\,\mathrm{d}x = x^2 + C$. Leaving the $+C$ off an indefinite integral is a marked error, not a stylistic choice.
Integrating xⁿ
Reversing the power rule gives the formula this whole topic rests on. Differentiating drops the power by one and multiplies by it, so integration does the opposite: raise the power by one, then divide by the new power.
The restriction $n \neq -1$ is there for a simple reason: if $n = -1$, then $n+1 = 0$, and the formula would divide by zero. Integrating $x^{-1}$ needs a different function altogether, which comes later in the course.
The formula only applies to a power of $x$. So convert first: $\sqrt{x}$ becomes $x^{1/2}$, $\dfrac{1}{x^2}$ becomes $x^{-2}$, and a fraction like $\dfrac{x^2+1}{x}$ splits into $x + x^{-1}$. Integration is also linear, so constant multiples come through unchanged and sums integrate term by term. Add one $C$ at the very end for the whole expression, not one per term.
Worked example 1
Differentiate your answer and you should get back what you started with: $\dfrac{d}{dx}\!\left(2x^3 - 2x^2 + 5x + C\right) = 6x^2 - 4x + 5$. That takes five seconds and catches almost every slip. The term worth watching is the constant $5$, which integrates to $5x$ rather than disappearing.
Worked example 2
Adding one to the power works identically whatever $n$ is: $\tfrac12$ becomes $\tfrac32$, and $-3$ becomes $-2$. The only power the formula cannot take is $n = -1$, where adding one gives zero and you would be dividing by it. Every other power of $x$ integrates by the same one-line rule.
Worked example 3
There is no product rule for integration at this level, so a bracket like this has to be multiplied out first. Once it is a sum of powers, integrate term by term.
The same move works on quotients. A fraction such as $\dfrac{x^2 + 1}{x}$ splits into $\dfrac{x^2}{x} + \dfrac{1}{x} = x + x^{-1}$, and each term is then a power you can handle. Whatever the expression looks like, get it into the form $kx^n + kx^n + \ldots$ before you start.
Two things account for most lost marks here. The first is forgetting $+C$. The second is dividing before raising the power: it is always add one to the power first, then divide by the power you now have. And do not try to integrate a product factor by factor — expand it into a sum first.
Check yourself
Find $\displaystyle\int \left(4x^3 - \dfrac{1}{x^2} + \sqrt{x}\right)\,\mathrm{d}x$.
(Hint: rewrite $\dfrac{1}{x^2}$ as $x^{-2}$ and $\sqrt{x}$ as $x^{1/2}$ first.)
Note what happened in step 3: $-x^{-2}$ integrated to $+x^{-1}$, because the two minus signs cancelled. A $\tfrac1x$ appears in the answer, but we never integrated $x^{-1}$ itself, so the excluded case did not arise.