Integration as the Reverse of Differentiation

Integration undoes differentiation. For the power rule that means:

$\int x^n \,dx = \frac{x^{n+1}}{n + 1} + c$ ($n \ne -1$), raise the power, divide by the new power, add $c$

The constant of integration is not pedantry: differentiating kills constants, so reversing must acknowledge the unknown one. The dropped $+c$ is the most predictable lost accuracy mark in 0606, weld it to the same pen stroke as the integral.

The standard forms

Each is “reverse the chain rule for a linear inside”, i.e. divide by the inside’s coefficient:

• $\int (ax + b)^n \,dx = \frac{(ax + b)^{n+1}}{a(n + 1)} + c$
• $\int e^{ax+b} \,dx = \frac{e^{ax+b}}{a} + c$
• $\int \sin(ax + b) \,dx = \frac{-\cos(ax + b)}{a} + c$ $\cdot$ $\int \cos(ax + b) \,dx = \frac{\sin(ax + b)}{a} + c$ (note the sign swap, integrating $\sin$ gives $-\cos$)

$\int (2x - 5)^4 \,dx = \frac{(2x - 5)^5}{10} + c$, divide by both the new power 5 and the inside coefficient 2.

Only linear insides work this way; 0606 doesn’t ask you to integrate through $(x^2 + 1)^3$, if you find yourself trying, re-read the question. Prepare awkward integrands first: expand brackets, split fractions ($\frac{x^3 + 2}{x^2} = x + 2x^{-2}$), rewrite roots as powers ($\sqrt{x} = x^{1/2}$).

Finding $c$: the curve-through-a-point question

A curve has $\frac{dy}{dx} = 3x^2 - 4$ and passes through $(2, 5)$. Find its equation. $y = x^3 - 4x + c$ (M, A, with the $c$) Through $(2, 5)$: $5 = 8 - 8 + c \to c = 5$ $y = x^3 - 4x + 5$

Integrate (with $c$), substitute the point, solve for $c$, state the full equation. Without the $+c$ this question cannot be answered, the structural reason the constant matters. The same find-the-constant logic powers kinematics initial conditions.

A self-check worth its ten seconds: differentiate your answer, it should reproduce the integrand exactly.

Common mistakes

• $+c$ dropped (indefinite integrals only, definite ones cancel it)
• The inside coefficient not divided out in the standard forms
• $\sin$/$\cos$ integration signs swapped
• Quotients integrated term-by-term without splitting first
• The found c never assembled back into a stated equation

Full topic context: Calculus notes · the exam drill: integration & area technique.