The toolkit for harder integrals
The normalization integral for the quantum harmonic oscillator ground state:
$$\int_{-\infty}^{\infty} x^2 e^{-\alpha x^2} \, dx = \, ?$$
This isn't a simple substitution. The $x^2$ and $e^{-\alpha x^2}$ are tangled together.
You need integration by parts.
Or consider consecutive reactions A → B → C requiring partial fractions decomposition.
You need partial fractions.
| Integrand looks like... | Technique |
|---|---|
| Product of two different types | Integration by parts |
| (polynomial) × (exp, sin, cos, ln) | Integration by parts |
| Rational function P(x)/Q(x) | Partial fractions |
| √(a² - x²) | Trig sub: x = a sin θ |
| √(a² + x²) | Trig sub: x = a tan θ |
| √(x² - a²) | Trig sub: x = a sec θ |
| Powers of sin and cos | Trig identities |
From the product rule: $(uv)' = u'v + uv'$
Rearrange and integrate:
$$\int u \, dv = uv - \int v \, du$$
Choose u in this priority order (earlier = better choice for u):
Identify: x (algebraic) × eˣ (exponential)
By LIATE: A comes before E → u = x, dv = eˣ dx
| u = x | dv = eˣ dx |
| du = dx | v = eˣ |
$$\int x e^x dx = xe^x - \int e^x dx = xe^x - e^x + C = e^x(x-1) + C$$
For integrals like $\int x^n e^x dx$, use a table to organize repeated by parts:
| Derivatives (u) | Integrals (dv) | Sign |
|---|---|---|
| x³ | eˣ | + |
| 3x² | eˣ | − |
| 6x | eˣ | + |
| 6 | eˣ | − |
| 0 | eˣ | + |
Result: $e^x(x^3 - 3x^2 + 6x - 6) + C$
Multiply diagonally, alternating signs!
Clever trick: write as ln x · 1
u = ln x, dv = dx → du = (1/x)dx, v = x
$$\int \ln x \, dx = x\ln x - \int x \cdot \frac{1}{x} dx = x\ln x - x + C = x(\ln x - 1) + C$$
For rational functions $\frac{P(x)}{Q(x)}$ where degree(P) < degree(Q).
Factor the denominator, then decompose into simpler fractions.
Each simple fraction is easy to integrate.
$$\frac{1}{(x-a)(x-b)} = \frac{A}{x-a} + \frac{B}{x-b}$$
$(x-a)^n$ in denominator requires:
$$\frac{A_1}{x-a} + \frac{A_2}{(x-a)^2} + \cdots + \frac{A_n}{(x-a)^n}$$
$(x^2 + bx + c)$ in denominator:
$$\frac{Ax + B}{x^2 + bx + c}$$
Split into two parts:
$$= \int \frac{2x}{x^2+1} dx + \int \frac{3}{x^2+1} dx$$
$$= \ln(x^2+1) + 3\arctan(x) + C$$
For integrands containing square roots of quadratics:
| Expression | Substitution | Identity used |
|---|---|---|
| √(a² - x²) | x = a sin θ | 1 - sin²θ = cos²θ |
| √(a² + x²) | x = a tan θ | 1 + tan²θ = sec²θ |
| √(x² - a²) | x = a sec θ | sec²θ - 1 = tan²θ |
Let x = a tan θ, dx = a sec²θ dθ
x² + a² = a²tan²θ + a² = a²sec²θ
$$\int \frac{1}{x^2 + a^2} dx = \int \frac{a\sec^2\theta}{a^2\sec^2\theta} d\theta = \frac{1}{a}\int d\theta = \frac{1}{a}\arctan\left(\frac{x}{a}\right) + C$$
For 2A → products with $-\frac{d[A]}{dt} = k[A]^2$:
$$\int \frac{d[A]}{[A]^2} = -k\int dt$$
$$-\frac{1}{[A]} = -kt + C$$
At t = 0: $C = -\frac{1}{[A]_0}$
$$\frac{1}{[A]} = \frac{1}{[A]_0} + kt$$
Half-life: $t_{1/2} = \frac{1}{k[A]_0}$ — depends on initial concentration!
The integral $\int x^2 e^{-\alpha x^2} dx$ appears in quantum mechanics for expectation values.
Let u = x, dv = xe^(-αx²)dx → v = -1/(2α)e^(-αx²)
$$\int_0^{\infty} x^2 e^{-\alpha x^2} dx = \left[-\frac{x}{2\alpha}e^{-\alpha x^2}\right]_0^{\infty} + \frac{1}{2\alpha}\int_0^{\infty} e^{-\alpha x^2} dx$$
Boundary terms vanish. Using $\int_0^{\infty} e^{-\alpha x^2} dx = \frac{1}{2}\sqrt{\frac{\pi}{\alpha}}$:
$$= \frac{1}{2\alpha} \cdot \frac{1}{2}\sqrt{\frac{\pi}{\alpha}} = \frac{\sqrt{\pi}}{4\alpha^{3/2}}$$
| Integral | Result |
|---|---|
| ∫ xeˣ dx | eˣ(x-1) + C |
| ∫ x sin x dx | -x cos x + sin x + C |
| ∫ ln x dx | x(ln x - 1) + C |
| ∫ 1/(x²+a²) dx | (1/a) arctan(x/a) + C |
| ∫ 1/((x-a)(x-b)) dx | (1/(a-b)) ln|(x-a)/(x-b)| + C |