Integration (Pure 3)
New integration methods
- A new standard integral: $\displaystyle\int \frac{1}{x^2 + a^2}\,dx = \frac{1}{a}\tan^{-1}\frac{x}{a} + C$.
- Partial fractions: split, then integrate each piece as a logarithm.
- The pattern $\dfrac{k\,f'(x)}{f(x)}$ → $k\ln|f(x)| + C$, e.g. $\displaystyle\int \frac{2x}{x^2+1}\,dx = \ln(x^2+1) + C$.
Practice
What is ∫ 2x/(x²+1) dx?
The top is the derivative of the bottom (k f′/f form), so the integral is ln(x²+1) + C.
Practice
∫ 1/(x²+a²) dx = (1/a) tan⁻¹(x/a) + C.
This is a standard Pure 3 integral linked to the inverse-tangent derivative.
By parts & by substitution
- Integration by parts (for a product): $\displaystyle\int u\,\frac{dv}{dx}\,dx = uv - \int v\,\frac{du}{dx}\,dx$.
- Integration by substitution: a change of variable turns a hard integral into an easy one.
- Example: $\displaystyle\int x\cos x\,dx = x\sin x + \cos x + C$ (by parts).
Practice
The integration-by-parts formula is ∫ u dv =:
Integration by parts: ∫ u dv = uv − ∫ v du.
You've got it
Key idea
- $\displaystyle\int \frac{1}{x^2+a^2}dx = \frac1a\tan^{-1}\frac{x}{a} + C$; $\displaystyle\int\frac{k f'(x)}{f(x)}dx = k\ln|f(x)| + C$
- by parts: $\int u\,dv = uv - \int v\,du$ (for products)
- by substitution: change the variable to simplify