Basic Integration (AA HL)
Integration undoes differentiation, and most of the time that means recognising which standard result applies. This page covers integrating powers of \(x\), the standard functions, linear composites via the reverse chain rule, and when to reach for integration by parts instead. It's part of the broader Integration topic.
32 questions on this sub-topic.
Key formulas
Covered under IB syllabus reference SL5.10: the indefinite integral of \(x^n\) (\(n\neq-1\)), \(\sin x\), \(\cos x\), \(e^x\) and \(\dfrac1x\), plus composites of these with a linear function \(ax+b\), by inspection or substitution.
Power rule
\(\displaystyle\int x^n\,dx = \dfrac{x^{n+1}}{n+1}+C,\ n\neq-1\)
✓ In the formula bookletReverse chain rule / substitution
Replace part of the integrand with a new variable \(u\) to turn an awkward integral into a standard one. For a straightforward linear composite like \(\cos(ax+b)\), it's usually faster to integrate directly and divide by \(a\).
Not a listed formula - techniqueIntegration by parts
\(\displaystyle\int u\,\dfrac{dv}{dx}\,dx = uv - \int v\,\dfrac{du}{dx}\,dx\)
Needed for products where one factor simplifies on differentiating, such as \(x\ln x\) or \(xe^{2x}\). Covered in full on the Integration by Parts page.
✓ In the formula bookletNeed the full syllabus wording and formula-booklet reference table, or a GDC walkthrough? See Integration.
Worked examples
Find \(\displaystyle\int\left(\dfrac{3}{x^2}+\sqrt{x}\right)dx.\)
Worked solution
Rewrite: \(3x^{-2}+x^{1/2}.\) M1
\(\int = -3x^{-1}+\tfrac{2}{3}x^{3/2}+C\) M1
\(= -\dfrac{3}{x}+\dfrac{2}{3}x^{3/2}+C.\) A1
Find \(\displaystyle\int \sin^2 x\,dx.\)
Worked solution
\(\sin^2 x = \dfrac{1 - \cos 2x}{2}.\) M1 A1
\(\int \dfrac{1 - \cos 2x}{2}\,dx\) M1 \(= \tfrac12 x - \tfrac14\sin 2x + C.\) A1 A1
Solve \(\dfrac{d^2y}{dx^2}=12x\) given \(y(0)=1\) and \(y'(0)=0.\)
Worked solution
\(y' = 6x^2 + C_1\); \(y'(0)=0 \Rightarrow C_1\) M1
\(= 0.\) A1
\(y = 2x^3 + C_2\); \(y(0)=1 \Rightarrow C_2\) M1
\(= 1.\) A1
\(y = 2x^3 + 1.\) A1
A curve has gradient \(\dfrac{dy}{dx} = 3x^2 - 6x\) and passes through \((2, 1).\)
(a) Find \(y\) in terms of \(x.\)
(b) Find \(y\) when \(x = 0.\)
Worked solution
(a) \(y = \int (3x^2 - 6x)\,dx\) M1
\(= x^3 - 3x^2 + C.\) A1
Using \((2,1)\): \(1 = 8 - 12 + C \Rightarrow C = 5.\) M1
\(y = x^3 - 3x^2 + 5.\) A1
(b) \(y(0) = 5.\) A1
Common mistakes
- Choosing \(u\) and \(dv\) poorly in integration by parts. Pick \(u\) to be the factor that gets simpler when differentiated (like \(x\) or \(\ln x\)) - the other way round usually makes the resulting integral harder, not easier.
- Forgetting to divide by the coefficient in a linear composite. \(\displaystyle\int \cos(3x+1)\,dx = \dfrac13\sin(3x+1)+C\), not \(\sin(3x+1)+C\) - the reverse chain rule always needs that extra division.
- Leaving a root or fraction unconverted to power form. \(\sqrt{x}\) and \(\dfrac{1}{x^2}\) need to become \(x^{1/2}\) and \(x^{-2}\) before the power rule can be applied - trying to integrate them in their original form leads nowhere.
Ready to practise properly?
32 basic integration questions, marked instantly like the real exam.
Quick answers
What is the reverse chain rule in integration?
It integrates a composite function of the form \(ax+b\) by inspection: integrate as normal, then divide by the coefficient of \(x\). For example, \(\int\cos(3x+1)\,dx = \tfrac13\sin(3x+1)+C\).
When do I use integration by parts instead of a simple rule?
Use integration by parts when integrating a product of two functions where one factor gets simpler when differentiated, such as \(x\ln x\) or \(xe^{2x}\). Pick \(u\) to be that simplifying factor.