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Mathematics · Ch 10 — Indefinite Integration

Elementary Integration Formulae

10.1.1

Elementary Integration Formulae

Every differentiation formula can be read in reverse to produce an integration formula. If ddxg(x)=f(x)\frac{d}{dx}g(x)=f(x), then automatically ∫f(x) dx=g(x)+c\int f(x)\,dx=g(x)+c. Applying this idea to each of the standard derivatives gives the list of elementary integrals recorded in the table above.

Take the power rule as the first illustration. Since ddx(xn+1n+1)=(n+1)xnn+1=xn\frac{d}{dx}\left(\frac{x^{n+1}}{n+1}\right)=\frac{(n+1)x^n}{n+1}=x^n (for n≠−1n\neq-1), reversing it gives ∫xn dx=xn+1n+1+c\int x^n\,dx=\frac{x^{n+1}}{n+1}+c. The same computation carried out for (ax+b)n(ax+b)^n in place of xnx^n — using ddx(ax+b)n+1(n+1)a=(ax+b)n\frac{d}{dx}\frac{(ax+b)^{n+1}}{(n+1)a}=(ax+b)^n — gives the more general ∫(ax+b)n dx=(ax+b)n+1n+1⋅1a+c\int (ax+b)^n\,dx=\frac{(ax+b)^{n+1}}{n+1}\cdot\frac1a+c. This "ax+bax+b" pattern repeats for every entry in the table: whenever we know ∫f(x) dx=g(x)+c\int f(x)\,dx=g(x)+c, replacing xx by ax+bax+b everywhere and dividing by aa gives ∫f(ax+b) dx=g(ax+b)⋅1a+c\int f(ax+b)\,dx=g(ax+b)\cdot\frac1a+c (this is proved formally as Corollary I of the substitution theorem in section 3.2.1). …

Table 1Standard elementary integrals

(i) ∫xn dx=xn+1n+1+c, n≠−1\int x^n\,dx = \dfrac{x^{n+1}}{n+1}+c,\ n\neq -1; ∫(ax+b)n dx=(ax+b)n+1n+1⋅1a+c\int (ax+b)^n\,dx = \dfrac{(ax+b)^{n+1}}{n+1}\cdot\dfrac1a+c (extends to rational n=p/qn=p/q).\n(ii) ∫ax dx=axlog⁡a+c, a>0\int a^x\,dx = \dfrac{a^x}{\log a}+c,\ a>0; ∫Aax+b dx=Aax+blog⁡A⋅1a+c, A>0\int A^{ax+b}\,dx=\dfrac{A^{ax+b}}{\log A}\cdot\dfrac1a+c,\ A>0.\n(iii) ∫ex dx=ex+c\int e^x\,dx=e^x+c; ∫eax+b dx=eax+b⋅1a+c\int e^{ax+b}\,dx=e^{ax+b}\cdot\dfrac1a+c.\n(iv) ∫cos⁡x dx=sin⁡x+c\int \cos x\,dx=\sin x+c; ∫cos⁡(ax+b) dx=sin⁡(ax+b)⋅1a+c\int \cos(ax+b)\,dx=\sin(ax+b)\cdot\dfrac1a+c.\n(v) ∫sin⁡x dx=−cos⁡x+c\int \sin x\,dx=-\cos x+c; ∫sin⁡(ax+b) dx=−cos⁡(ax+b)⋅1a+c\int \sin(ax+b)\,dx=-\cos(ax+b)\cdot\dfrac1a+c.\n(vi) ∫sec⁡2x dx=tan⁡x+c\int \sec^2x\,dx=\tan x+c; ∫sec⁡2(ax+b) dx=tan⁡(ax+b)⋅1a+c\int \sec^2(ax+b)\,dx=\tan(ax+b)\cdot\dfrac1a+c.\n(vii) ∫sec⁡xtan⁡x dx=sec⁡x+c\int \sec x\tan x\,dx=\sec x+c; ∫sec⁡(ax+b)tan⁡(ax+b) dx=sec⁡(ax+b)⋅1a+c\int \sec(ax+b)\tan(ax+b)\,dx=\sec(ax+b)\cdot\dfrac1a+c.\n(viii) ∫csc⁡xcot⁡x dx=−csc⁡x+c\int \csc x\cot x\,dx=-\csc x+c; ∫csc⁡(ax+b)cot⁡(ax+b) dx=−csc⁡(ax+b)⋅1a+c\int \csc(ax+b)\cot(ax+b)\,dx=-\csc(ax+b)\cdot\dfrac1a+c.\n(ix) ∫csc⁡2x dx=−cot⁡x+c\int \csc^2x\,dx=-\cot x+c; $ …