Why Integration Exists: Adding Up A Rate To Get A Total
Differentiation turns a quantity into its rate of change.
Integration runs that backwards: from a rate, it rebuilds the total.
That is exactly what real problems need: from a velocity you want the distance travelled, from a rate of water flow you want the volume collected, and from a curve you want the area underneath it.
At SL you reversed the power rule for whole-number powers. Here you reverse the full derivative table from 5.9, and you learn to undo the chain rule so you can integrate composite functions.
Analogy
If differentiating is unpacking a total into its moment-by-moment rate, integrating is packing those moments back up into a running total.
The '+C' is the piece of information packing loses: many totals share the same rate, differing only by a starting amount.
The Standard Integrals: Reversing The Derivative Table
Each standard derivative read backwards gives a standard integral, all of which are in the formula booklet.
$$\displaystyle\int x^n\,dx = \dfrac{x^{n+1}}{n+1} + C$$ which is valid for any rational $n$ except $n = -1$.
$$\displaystyle\int \dfrac{1}{x}\,dx = \ln|x| + C$$ the special case that fills the hole at $n = -1$.
$$\displaystyle\int \sin x\,dx = -\cos x + C$$
$$\displaystyle\int \cos x\,dx = \sin x + C$$
$$\displaystyle\int \dfrac{1}{\cos^2 x}\,dx = \tan x + C$$
$$\displaystyle\int e^x\,dx = e^x + C$
Hint
Why $n = -1$ is excluded: the power-rule formula would divide by $n+1 = 0$, so that single case is handled by the logarithm instead.
Note
The '+C' (constant of integration) is compulsory for every indefinite integral.
A boundary condition, such as a known value $f(a) = b$, pins down $C$ to give one particular function.
Trig integrals assume $x$ is in radians, matching the derivative table.
Pinning Down The Constant
If you know the rate and one data point, you can recover the exact total.
Example question
A curve has gradient $\dfrac{dy}{dx} = 3x^2 + 2$ and passes through $(1, 5)$. Find $y$ in terms of $x$.
Solution
Integrate: $y = x^3 + 2x + C$.
Use the point $(1,5)$: $1 + 2 + C = 5$, so $C = 2$.
Therefore $y = x^3 + 2x + 2$.
The Reverse Chain Rule: Integrating By Inspection
The chain rule says differentiating a composite multiplies by the derivative of the inside.
Reading that backwards lets you integrate many composites at sight. $$ \int f'(g(x))\,g'(x)\,dx = f(g(x)) + C $$
The most common exam version is a linear inside, $ax+b$, which introduces a $\dfrac{1}{a}$ factor when you reverse it.
$\displaystyle\int e^{ax+b}\,dx = \dfrac{1}{a}e^{ax+b} + C$, for example $\displaystyle\int e^{2x}\,dx = \dfrac{1}{2}e^{2x} + C$.
$\displaystyle\int \cos(3x)\,dx = \dfrac{1}{3}\sin(3x) + C$, dividing by the $3$ from the inside.
Worked example, $\displaystyle\int 2x\,e^{x^2}\,dx$: the $2x$ is exactly the derivative of the inside $x^2$, so the integral is $e^{x^2} + C$.
Common Mistake
When the inside is $ax+b$, do not forget to divide by $a$.
$\displaystyle\int \cos(3x)\,dx$ is $\dfrac{1}{3}\sin(3x) + C$, not $\sin(3x) + C$.
Example question
Find $\displaystyle\int \dfrac{x}{x^2+1}\,dx$.
Solution
Notice the top is almost the derivative of the bottom: $$\dfrac{d}{dx}(x^2+1) = 2x$$
Adjust for the missing factor of 2: $$\displaystyle\int \dfrac{x}{x^2+1}\,dx = \dfrac{1}{2}\int \dfrac{2x}{x^2+1}\,dx$$
This reverses to a logarithm: $$\dfrac{1}{2}\ln(x^2+1) + C$$
Integration By Substitution: A Written Method For The Reverse Chain
When you cannot spot the reverse chain by eye, substitution makes it systematic by renaming the inside function.
The method, step by step:
Let $u$ be the inside function, then find $\dfrac{du}{dx}$ and write $du = g'(x)\,dx$.
Replace every $x$-part of the integral with $u$-parts, so the integral is entirely in $u$.
Integrate in $u$, then substitute back to $x$ (for an indefinite integral).
For a definite integral you may instead change the limits to $u$-values, which avoids substituting back.
Example
For $\displaystyle\int 2x(x^2+1)^3\,dx$:
Let $u = x^2+1$, so $du = 2x\,dx$.
The integral becomes $\displaystyle\int u^3\,du = \dfrac{u^4}{4} + C = \dfrac{(x^2+1)^4}{4} + C$.
Exam technique
The standard integrals are in the formula booklet, but substitution is a method, so show the $u$ and $du$ lines to earn the method marks.
On a definite integral, every AI paper lets you get the numerical value straight from the GDC, so write the integral expression, then evaluate on the calculator.
If you change the variable to $u$, either change the limits to match or substitute back to $x$ before using the original limits; never mix the two.
Example question
Evaluate $\displaystyle\int_0^1 2x(x^2+1)^3\,dx$.
Solution
Substitute $u = x^2+1$, $du = 2x\,dx$.
Limits become $u = 1$ (at $x=0$) and $u = 2$ (at $x=1$).
The integral is $\displaystyle\int_1^2 u^3\,du = \left[\dfrac{u^4}{4}\right]_1^2 = \dfrac{16 - 1}{4}$.
So the value is $\dfrac{15}{4} = 3.75$, which the GDC confirms directly.
Active recall
Write the standard integral of $x^n$ and state the one value of $n$ it excludes, and why.
What is $\displaystyle\int \dfrac{1}{x}\,dx$, and why is it needed?
When integrating $\cos(ax+b)$, what extra factor appears and where does it come from?
List the three steps of integration by substitution.
For a definite integral solved by substitution, what are the two valid ways to handle the limits?
